How to find dimension of eigenspace?
up vote
0
down vote
favorite
Given $lambda$ = 2 and matrix A:
A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$
My textbook says that to find the eigenvectors of $lambda = 2$, solve the homogeneous linear system represented by $(2I - A)x = 0$.
$2I - A$ =
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
"This implies that $x_2$ = 0".
How is the matrix
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
computed and how does this imply that $x_2$ = 0? My textbook does not give an explanation.
linear-algebra
asked Dec 4 at 20:18
Evan Kim
758
|
show 3 more comments
up vote
0
down vote
favorite
Given $lambda$ = 2 and matrix A:
A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$
My textbook says that to find the eigenvectors of $lambda = 2$, solve the homogeneous linear system represented by $(2I - A)x = 0$.
$2I - A$ =
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
"This implies that $x_2$ = 0".
How is the matrix
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
computed and how does this imply that $x_2$ = 0? My textbook does not give an explanation.
linear-algebra
asked Dec 4 at 20:18
Evan Kim
758
The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
– Michael Burr
Dec 4 at 20:22
The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
– Bernard
Dec 4 at 20:27
@Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
– Federico
Dec 4 at 20:27
$dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
– Federico
Dec 4 at 20:29
@Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
– Evan Kim
Dec 4 at 21:25
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given $lambda$ = 2 and matrix A:
A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$
My textbook says that to find the eigenvectors of $lambda = 2$, solve the homogeneous linear system represented by $(2I - A)x = 0$.
$2I - A$ =
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
"This implies that $x_2$ = 0".
How is the matrix
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
computed and how does this imply that $x_2$ = 0? My textbook does not give an explanation.
linear-algebra
asked Dec 4 at 20:18
Evan Kim
758
Given $lambda$ = 2 and matrix A:
A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$
My textbook says that to find the eigenvectors of $lambda = 2$, solve the homogeneous linear system represented by $(2I - A)x = 0$.
$2I - A$ =
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
"This implies that $x_2$ = 0".
How is the matrix
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
computed and how does this imply that $x_2$ = 0? My textbook does not give an explanation.
linear-algebra
linear-algebra
asked Dec 4 at 20:18
Evan Kim
758
asked Dec 4 at 20:18
Evan Kim
758
asked Dec 4 at 20:18
Evan Kim
758
asked Dec 4 at 20:18
Evan Kim
758
asked Dec 4 at 20:18
Evan Kim
758
758
The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
– Michael Burr
Dec 4 at 20:22
The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
– Bernard
Dec 4 at 20:27
@Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
– Federico
Dec 4 at 20:27
$dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
– Federico
Dec 4 at 20:29
@Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
– Evan Kim
Dec 4 at 21:25
|
show 3 more comments
The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
– Michael Burr
Dec 4 at 20:22
The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
– Bernard
Dec 4 at 20:27
@Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
– Federico
Dec 4 at 20:27
$dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
– Federico
Dec 4 at 20:29
@Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
– Evan Kim
Dec 4 at 21:25
The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
– Michael Burr
Dec 4 at 20:22
The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
– Michael Burr
Dec 4 at 20:22
The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
– Bernard
Dec 4 at 20:27
The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
– Bernard
Dec 4 at 20:27
@Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
– Federico
Dec 4 at 20:27
@Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
– Federico
Dec 4 at 20:27
$dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
– Federico
Dec 4 at 20:29
$dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
– Federico
Dec 4 at 20:29
@Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
– Evan Kim
Dec 4 at 21:25
@Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
– Evan Kim
Dec 4 at 21:25
|
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
The matrix $2I-A$ is simply computed as $$begin{bmatrix}2 & 0 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}-begin{bmatrix}2 & 1 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}=begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$$
If a vector $x$ with coordinates $x_1,x_2,x_3$ satisfies $(2I-A)x=0$, then it must be that $x_2=0$. This is because $$begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}implies begin{bmatrix}-x_2\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$ so we must have $x_2=0$. Notice that $x_1$ and $x_3$ can be any number, so what does this tell you about the dimension of the eigenspace (i.e. the dimension of ${xinmathbb C^3:(2A-I)x=0}$)?
answered Dec 4 at 20:31
Dave
8,60611033
add a comment |
up vote
2
down vote
We have that
$$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}vec x=vec 0 implies vec x=(x_1,0,x_3)$$
therefore what is the dimension of the eigenspace?
answered Dec 4 at 20:20
gimusi
92.5k94495
add a comment |
up vote
0
down vote
You must find all a basis for the space of vectors such that:
$Amathbf v = lambda mathbf v$
and with a little algebra
$(A- lambda I)mathbf v = mathbf 0$
If $x_2 ne 0$
$(A- 2I)mathbf v = begin{bmatrix} x_2\0\0 end{bmatrix} ne mathbf 0$
answered Dec 4 at 20:39
Doug M
43.5k31854
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026100%2fhow-to-find-dimension-of-eigenspace%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The matrix $2I-A$ is simply computed as $$begin{bmatrix}2 & 0 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}-begin{bmatrix}2 & 1 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}=begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$$
If a vector $x$ with coordinates $x_1,x_2,x_3$ satisfies $(2I-A)x=0$, then it must be that $x_2=0$. This is because $$begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}implies begin{bmatrix}-x_2\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$ so we must have $x_2=0$. Notice that $x_1$ and $x_3$ can be any number, so what does this tell you about the dimension of the eigenspace (i.e. the dimension of ${xinmathbb C^3:(2A-I)x=0}$)?
answered Dec 4 at 20:31
Dave
8,60611033
add a comment |
up vote
2
down vote
accepted
The matrix $2I-A$ is simply computed as $$begin{bmatrix}2 & 0 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}-begin{bmatrix}2 & 1 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}=begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$$
If a vector $x$ with coordinates $x_1,x_2,x_3$ satisfies $(2I-A)x=0$, then it must be that $x_2=0$. This is because $$begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}implies begin{bmatrix}-x_2\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$ so we must have $x_2=0$. Notice that $x_1$ and $x_3$ can be any number, so what does this tell you about the dimension of the eigenspace (i.e. the dimension of ${xinmathbb C^3:(2A-I)x=0}$)?
answered Dec 4 at 20:31
Dave
8,60611033
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The matrix $2I-A$ is simply computed as $$begin{bmatrix}2 & 0 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}-begin{bmatrix}2 & 1 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}=begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$$
If a vector $x$ with coordinates $x_1,x_2,x_3$ satisfies $(2I-A)x=0$, then it must be that $x_2=0$. This is because $$begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}implies begin{bmatrix}-x_2\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$ so we must have $x_2=0$. Notice that $x_1$ and $x_3$ can be any number, so what does this tell you about the dimension of the eigenspace (i.e. the dimension of ${xinmathbb C^3:(2A-I)x=0}$)?
answered Dec 4 at 20:31
Dave
8,60611033
The matrix $2I-A$ is simply computed as $$begin{bmatrix}2 & 0 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}-begin{bmatrix}2 & 1 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}=begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$$
If a vector $x$ with coordinates $x_1,x_2,x_3$ satisfies $(2I-A)x=0$, then it must be that $x_2=0$. This is because $$begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}implies begin{bmatrix}-x_2\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$ so we must have $x_2=0$. Notice that $x_1$ and $x_3$ can be any number, so what does this tell you about the dimension of the eigenspace (i.e. the dimension of ${xinmathbb C^3:(2A-I)x=0}$)?
answered Dec 4 at 20:31
Dave
8,60611033
answered Dec 4 at 20:31
Dave
8,60611033
answered Dec 4 at 20:31
Dave
8,60611033
answered Dec 4 at 20:31
Dave
8,60611033
8,60611033
add a comment |
add a comment |
up vote
2
down vote
We have that
$$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}vec x=vec 0 implies vec x=(x_1,0,x_3)$$
therefore what is the dimension of the eigenspace?
answered Dec 4 at 20:20
gimusi
92.5k94495
add a comment |
up vote
2
down vote
We have that
$$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}vec x=vec 0 implies vec x=(x_1,0,x_3)$$
therefore what is the dimension of the eigenspace?
answered Dec 4 at 20:20
gimusi
92.5k94495
add a comment |
up vote
2
down vote
up vote
2
down vote
We have that
$$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}vec x=vec 0 implies vec x=(x_1,0,x_3)$$
therefore what is the dimension of the eigenspace?
answered Dec 4 at 20:20
gimusi
92.5k94495
We have that
$$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}vec x=vec 0 implies vec x=(x_1,0,x_3)$$
therefore what is the dimension of the eigenspace?
answered Dec 4 at 20:20
gimusi
92.5k94495
answered Dec 4 at 20:20
gimusi
92.5k94495
answered Dec 4 at 20:20
gimusi
92.5k94495
answered Dec 4 at 20:20
gimusi
92.5k94495
92.5k94495
add a comment |
add a comment |
up vote
0
down vote
You must find all a basis for the space of vectors such that:
$Amathbf v = lambda mathbf v$
and with a little algebra
$(A- lambda I)mathbf v = mathbf 0$
If $x_2 ne 0$
$(A- 2I)mathbf v = begin{bmatrix} x_2\0\0 end{bmatrix} ne mathbf 0$
answered Dec 4 at 20:39
Doug M
43.5k31854
add a comment |
up vote
0
down vote
You must find all a basis for the space of vectors such that:
$Amathbf v = lambda mathbf v$
and with a little algebra
$(A- lambda I)mathbf v = mathbf 0$
If $x_2 ne 0$
$(A- 2I)mathbf v = begin{bmatrix} x_2\0\0 end{bmatrix} ne mathbf 0$
answered Dec 4 at 20:39
Doug M
43.5k31854
add a comment |
up vote
0
down vote
up vote
0
down vote
You must find all a basis for the space of vectors such that:
$Amathbf v = lambda mathbf v$
and with a little algebra
$(A- lambda I)mathbf v = mathbf 0$
If $x_2 ne 0$
$(A- 2I)mathbf v = begin{bmatrix} x_2\0\0 end{bmatrix} ne mathbf 0$
answered Dec 4 at 20:39
Doug M
43.5k31854
You must find all a basis for the space of vectors such that:
$Amathbf v = lambda mathbf v$
and with a little algebra
$(A- lambda I)mathbf v = mathbf 0$
If $x_2 ne 0$
$(A- 2I)mathbf v = begin{bmatrix} x_2\0\0 end{bmatrix} ne mathbf 0$
answered Dec 4 at 20:39
Doug M
43.5k31854
answered Dec 4 at 20:39
Doug M
43.5k31854
answered Dec 4 at 20:39
Doug M
43.5k31854
answered Dec 4 at 20:39
Doug M
43.5k31854
43.5k31854
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026100%2fhow-to-find-dimension-of-eigenspace%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
– Michael Burr
Dec 4 at 20:22
The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
– Bernard
Dec 4 at 20:27
@Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
– Federico
Dec 4 at 20:27
$dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
– Federico
Dec 4 at 20:29
@Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
– Evan Kim
Dec 4 at 21:25