Quick question about eigenvalues of $3times3$ matrix
I'm looking over solutions from past exams. For this problem, the solution states “By inspection, we see that $2$ is an eigenvalue”. Given arbitrary $3times3$ matrix below$$begin{bmatrix}2&a&b \0&c&d\0&e&fend{bmatrix}$$
How is it obvious that $2$ is an eigenvalue? other constants are arbitrary so I can't assume it's an upper triangular matrix.
linear-algebra matrices eigenvalues-eigenvectors
edited Dec 12 '18 at 8:36
José Carlos Santos
152k22123226
asked Dec 12 '18 at 8:31
BradBrad
111
add a comment |
I'm looking over solutions from past exams. For this problem, the solution states “By inspection, we see that $2$ is an eigenvalue”. Given arbitrary $3times3$ matrix below$$begin{bmatrix}2&a&b \0&c&d\0&e&fend{bmatrix}$$
How is it obvious that $2$ is an eigenvalue? other constants are arbitrary so I can't assume it's an upper triangular matrix.
linear-algebra matrices eigenvalues-eigenvectors
edited Dec 12 '18 at 8:36
José Carlos Santos
152k22123226
asked Dec 12 '18 at 8:31
BradBrad
111
add a comment |
I'm looking over solutions from past exams. For this problem, the solution states “By inspection, we see that $2$ is an eigenvalue”. Given arbitrary $3times3$ matrix below$$begin{bmatrix}2&a&b \0&c&d\0&e&fend{bmatrix}$$
How is it obvious that $2$ is an eigenvalue? other constants are arbitrary so I can't assume it's an upper triangular matrix.
linear-algebra matrices eigenvalues-eigenvectors
edited Dec 12 '18 at 8:36
José Carlos Santos
152k22123226
asked Dec 12 '18 at 8:31
BradBrad
111
I'm looking over solutions from past exams. For this problem, the solution states “By inspection, we see that $2$ is an eigenvalue”. Given arbitrary $3times3$ matrix below$$begin{bmatrix}2&a&b \0&c&d\0&e&fend{bmatrix}$$
How is it obvious that $2$ is an eigenvalue? other constants are arbitrary so I can't assume it's an upper triangular matrix.
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
edited Dec 12 '18 at 8:36
José Carlos Santos
152k22123226
asked Dec 12 '18 at 8:31
BradBrad
111
edited Dec 12 '18 at 8:36
José Carlos Santos
152k22123226
asked Dec 12 '18 at 8:31
BradBrad
111
edited Dec 12 '18 at 8:36
José Carlos Santos
152k22123226
edited Dec 12 '18 at 8:36
José Carlos Santos
152k22123226
edited Dec 12 '18 at 8:36
José Carlos Santos
152k22123226
152k22123226
asked Dec 12 '18 at 8:31
BradBrad
111
asked Dec 12 '18 at 8:31
BradBrad
111
asked Dec 12 '18 at 8:31
BradBrad
111
111
add a comment |
add a comment |
5 Answers
5
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votes
It is obvious since, if $M$ is your matrix, then$$M.(1,0,0)=(2,0,0)=2(1,0,0).$$
answered Dec 12 '18 at 8:34
José Carlos SantosJosé Carlos Santos
152k22123226
add a comment |
Consider $det (A-lambda I)$ and expand the determinant through first column.
answered Dec 12 '18 at 8:34
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
add a comment |
It is obvious since, if you subtract $2I$, the matrix has a zero column. Thus if $M$ is your matrix,
$$
det(M-2I)=0
$$
answered Dec 12 '18 at 8:35
ArthurArthur
111k7105186
add a comment |
The columns of the matrix are the images of the basis vectors. The first column is a scalar multiple of $(1,0,0)^T$, therefore...
answered Dec 12 '18 at 10:55
amdamd
29.3k21050
add a comment |
This is a block upper triangular matrix of the form of $$M=begin{bmatrix} A & B \ O & Dend{bmatrix}$$, the eigenvalues of $M$ are eigenvalues of $A$ and $D$ since
$$det(M-lambda I)=det(A-lambda I_1) det(D-lambda I_2)$$
where $I_2$ and $I_2$ are identity matrices of appropriate sizes.
answered Dec 12 '18 at 8:35
Siong Thye GohSiong Thye Goh
99.6k1464117
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is obvious since, if $M$ is your matrix, then$$M.(1,0,0)=(2,0,0)=2(1,0,0).$$
answered Dec 12 '18 at 8:34
José Carlos SantosJosé Carlos Santos
152k22123226
add a comment |
It is obvious since, if $M$ is your matrix, then$$M.(1,0,0)=(2,0,0)=2(1,0,0).$$
answered Dec 12 '18 at 8:34
José Carlos SantosJosé Carlos Santos
152k22123226
add a comment |
It is obvious since, if $M$ is your matrix, then$$M.(1,0,0)=(2,0,0)=2(1,0,0).$$
answered Dec 12 '18 at 8:34
José Carlos SantosJosé Carlos Santos
152k22123226
It is obvious since, if $M$ is your matrix, then$$M.(1,0,0)=(2,0,0)=2(1,0,0).$$
answered Dec 12 '18 at 8:34
José Carlos SantosJosé Carlos Santos
152k22123226
answered Dec 12 '18 at 8:34
José Carlos SantosJosé Carlos Santos
152k22123226
answered Dec 12 '18 at 8:34
José Carlos SantosJosé Carlos Santos
152k22123226
answered Dec 12 '18 at 8:34
José Carlos SantosJosé Carlos Santos
152k22123226
152k22123226
add a comment |
add a comment |
Consider $det (A-lambda I)$ and expand the determinant through first column.
answered Dec 12 '18 at 8:34
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
add a comment |
Consider $det (A-lambda I)$ and expand the determinant through first column.
answered Dec 12 '18 at 8:34
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
add a comment |
Consider $det (A-lambda I)$ and expand the determinant through first column.
answered Dec 12 '18 at 8:34
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
Consider $det (A-lambda I)$ and expand the determinant through first column.
answered Dec 12 '18 at 8:34
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
answered Dec 12 '18 at 8:34
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
answered Dec 12 '18 at 8:34
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
answered Dec 12 '18 at 8:34
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
51.6k31955
add a comment |
add a comment |
It is obvious since, if you subtract $2I$, the matrix has a zero column. Thus if $M$ is your matrix,
$$
det(M-2I)=0
$$
answered Dec 12 '18 at 8:35
ArthurArthur
111k7105186
add a comment |
It is obvious since, if you subtract $2I$, the matrix has a zero column. Thus if $M$ is your matrix,
$$
det(M-2I)=0
$$
answered Dec 12 '18 at 8:35
ArthurArthur
111k7105186
add a comment |
It is obvious since, if you subtract $2I$, the matrix has a zero column. Thus if $M$ is your matrix,
$$
det(M-2I)=0
$$
answered Dec 12 '18 at 8:35
ArthurArthur
111k7105186
It is obvious since, if you subtract $2I$, the matrix has a zero column. Thus if $M$ is your matrix,
$$
det(M-2I)=0
$$
answered Dec 12 '18 at 8:35
ArthurArthur
111k7105186
answered Dec 12 '18 at 8:35
ArthurArthur
111k7105186
answered Dec 12 '18 at 8:35
ArthurArthur
111k7105186
answered Dec 12 '18 at 8:35
ArthurArthur
111k7105186
111k7105186
add a comment |
add a comment |
The columns of the matrix are the images of the basis vectors. The first column is a scalar multiple of $(1,0,0)^T$, therefore...
answered Dec 12 '18 at 10:55
amdamd
29.3k21050
add a comment |
The columns of the matrix are the images of the basis vectors. The first column is a scalar multiple of $(1,0,0)^T$, therefore...
answered Dec 12 '18 at 10:55
amdamd
29.3k21050
add a comment |
The columns of the matrix are the images of the basis vectors. The first column is a scalar multiple of $(1,0,0)^T$, therefore...
answered Dec 12 '18 at 10:55
amdamd
29.3k21050
The columns of the matrix are the images of the basis vectors. The first column is a scalar multiple of $(1,0,0)^T$, therefore...
answered Dec 12 '18 at 10:55
amdamd
29.3k21050
answered Dec 12 '18 at 10:55
amdamd
29.3k21050
answered Dec 12 '18 at 10:55
amdamd
29.3k21050
answered Dec 12 '18 at 10:55
amdamd
29.3k21050
29.3k21050
add a comment |
add a comment |
This is a block upper triangular matrix of the form of $$M=begin{bmatrix} A & B \ O & Dend{bmatrix}$$, the eigenvalues of $M$ are eigenvalues of $A$ and $D$ since
$$det(M-lambda I)=det(A-lambda I_1) det(D-lambda I_2)$$
where $I_2$ and $I_2$ are identity matrices of appropriate sizes.
answered Dec 12 '18 at 8:35
Siong Thye GohSiong Thye Goh
99.6k1464117
add a comment |
This is a block upper triangular matrix of the form of $$M=begin{bmatrix} A & B \ O & Dend{bmatrix}$$, the eigenvalues of $M$ are eigenvalues of $A$ and $D$ since
$$det(M-lambda I)=det(A-lambda I_1) det(D-lambda I_2)$$
where $I_2$ and $I_2$ are identity matrices of appropriate sizes.
answered Dec 12 '18 at 8:35
Siong Thye GohSiong Thye Goh
99.6k1464117
add a comment |
This is a block upper triangular matrix of the form of $$M=begin{bmatrix} A & B \ O & Dend{bmatrix}$$, the eigenvalues of $M$ are eigenvalues of $A$ and $D$ since
$$det(M-lambda I)=det(A-lambda I_1) det(D-lambda I_2)$$
where $I_2$ and $I_2$ are identity matrices of appropriate sizes.
answered Dec 12 '18 at 8:35
Siong Thye GohSiong Thye Goh
99.6k1464117
This is a block upper triangular matrix of the form of $$M=begin{bmatrix} A & B \ O & Dend{bmatrix}$$, the eigenvalues of $M$ are eigenvalues of $A$ and $D$ since
$$det(M-lambda I)=det(A-lambda I_1) det(D-lambda I_2)$$
where $I_2$ and $I_2$ are identity matrices of appropriate sizes.
answered Dec 12 '18 at 8:35
Siong Thye GohSiong Thye Goh
99.6k1464117
answered Dec 12 '18 at 8:35
Siong Thye GohSiong Thye Goh
99.6k1464117
answered Dec 12 '18 at 8:35
Siong Thye GohSiong Thye Goh
99.6k1464117
answered Dec 12 '18 at 8:35
Siong Thye GohSiong Thye Goh
99.6k1464117
99.6k1464117
add a comment |
add a comment |
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