Dimension of the null space of an invertible square matrix $A$
$begingroup$
Let $A$ be an $m times n$ matrix of real numbers and $cl(A)$ denotes the column space of $A$, $R(A)$ denotes the row space of $A$ and $N(A)$ denotes the Null space of $A$.
I want to prove this statement.
If $m=n$ and $det(A) neq 0$ then $dim(N(A))=0$
My attempt.
An $m times n$ matrix of real numbers is a linear transformation $T: mathbb{R}^n to mathbb{R}^m$
let $m=n=k implies$ $T: mathbb{R}^k to mathbb{R}^k$ Now since the matrix $A$ is invertible $implies$ $T$ has an inverse $T^{-1}$ $implies$ $T$ is injective thus the Null space $N(T)={0} implies $ $dim(N(T))=0$
linear-algebra matrices proof-verification linear-transformations
edited Dec 22 '18 at 7:21
Shubham Johri
5,082717
asked Dec 22 '18 at 7:03
Dreamer123Dreamer123
30429
$endgroup$
add a comment |
$begingroup$
Let $A$ be an $m times n$ matrix of real numbers and $cl(A)$ denotes the column space of $A$, $R(A)$ denotes the row space of $A$ and $N(A)$ denotes the Null space of $A$.
I want to prove this statement.
If $m=n$ and $det(A) neq 0$ then $dim(N(A))=0$
My attempt.
An $m times n$ matrix of real numbers is a linear transformation $T: mathbb{R}^n to mathbb{R}^m$
let $m=n=k implies$ $T: mathbb{R}^k to mathbb{R}^k$ Now since the matrix $A$ is invertible $implies$ $T$ has an inverse $T^{-1}$ $implies$ $T$ is injective thus the Null space $N(T)={0} implies $ $dim(N(T))=0$
linear-algebra matrices proof-verification linear-transformations
edited Dec 22 '18 at 7:21
Shubham Johri
5,082717
asked Dec 22 '18 at 7:03
Dreamer123Dreamer123
30429
$endgroup$
1
$begingroup$
Your proof is correct. Was that your question?
$endgroup$
– Jonas
Dec 22 '18 at 7:08
$begingroup$
I am always suspicious if there is a mistake :)
$endgroup$
– Dreamer123
Dec 22 '18 at 7:12
1
$begingroup$
In future, use theproof-verificationtag
$endgroup$
– Shubham Johri
Dec 22 '18 at 7:20
add a comment |
$begingroup$
Let $A$ be an $m times n$ matrix of real numbers and $cl(A)$ denotes the column space of $A$, $R(A)$ denotes the row space of $A$ and $N(A)$ denotes the Null space of $A$.
I want to prove this statement.
If $m=n$ and $det(A) neq 0$ then $dim(N(A))=0$
My attempt.
An $m times n$ matrix of real numbers is a linear transformation $T: mathbb{R}^n to mathbb{R}^m$
let $m=n=k implies$ $T: mathbb{R}^k to mathbb{R}^k$ Now since the matrix $A$ is invertible $implies$ $T$ has an inverse $T^{-1}$ $implies$ $T$ is injective thus the Null space $N(T)={0} implies $ $dim(N(T))=0$
linear-algebra matrices proof-verification linear-transformations
edited Dec 22 '18 at 7:21
Shubham Johri
5,082717
asked Dec 22 '18 at 7:03
Dreamer123Dreamer123
30429
$endgroup$
Let $A$ be an $m times n$ matrix of real numbers and $cl(A)$ denotes the column space of $A$, $R(A)$ denotes the row space of $A$ and $N(A)$ denotes the Null space of $A$.
I want to prove this statement.
If $m=n$ and $det(A) neq 0$ then $dim(N(A))=0$
My attempt.
An $m times n$ matrix of real numbers is a linear transformation $T: mathbb{R}^n to mathbb{R}^m$
let $m=n=k implies$ $T: mathbb{R}^k to mathbb{R}^k$ Now since the matrix $A$ is invertible $implies$ $T$ has an inverse $T^{-1}$ $implies$ $T$ is injective thus the Null space $N(T)={0} implies $ $dim(N(T))=0$
linear-algebra matrices proof-verification linear-transformations
linear-algebra matrices proof-verification linear-transformations
edited Dec 22 '18 at 7:21
Shubham Johri
5,082717
asked Dec 22 '18 at 7:03
Dreamer123Dreamer123
30429
edited Dec 22 '18 at 7:21
Shubham Johri
5,082717
asked Dec 22 '18 at 7:03
Dreamer123Dreamer123
30429
edited Dec 22 '18 at 7:21
Shubham Johri
5,082717
edited Dec 22 '18 at 7:21
Shubham Johri
5,082717
edited Dec 22 '18 at 7:21
Shubham Johri
5,082717
5,082717
asked Dec 22 '18 at 7:03
Dreamer123Dreamer123
30429
asked Dec 22 '18 at 7:03
Dreamer123Dreamer123
30429
asked Dec 22 '18 at 7:03
Dreamer123Dreamer123
30429
30429
1
$begingroup$
Your proof is correct. Was that your question?
$endgroup$
– Jonas
Dec 22 '18 at 7:08
$begingroup$
I am always suspicious if there is a mistake :)
$endgroup$
– Dreamer123
Dec 22 '18 at 7:12
1
$begingroup$
In future, use theproof-verificationtag
$endgroup$
– Shubham Johri
Dec 22 '18 at 7:20
add a comment |
1
$begingroup$
Your proof is correct. Was that your question?
$endgroup$
– Jonas
Dec 22 '18 at 7:08
$begingroup$
I am always suspicious if there is a mistake :)
$endgroup$
– Dreamer123
Dec 22 '18 at 7:12
1
$begingroup$
In future, use theproof-verificationtag
$endgroup$
– Shubham Johri
Dec 22 '18 at 7:20
1
1
$begingroup$
Your proof is correct. Was that your question?
$endgroup$
– Jonas
Dec 22 '18 at 7:08
$begingroup$
Your proof is correct. Was that your question?
$endgroup$
– Jonas
Dec 22 '18 at 7:08
$begingroup$
I am always suspicious if there is a mistake :)
$endgroup$
– Dreamer123
Dec 22 '18 at 7:12
$begingroup$
I am always suspicious if there is a mistake :)
$endgroup$
– Dreamer123
Dec 22 '18 at 7:12
1
1
$begingroup$
In future, use the
proof-verification tag$endgroup$
– Shubham Johri
Dec 22 '18 at 7:20
$begingroup$
In future, use the
proof-verification tag$endgroup$
– Shubham Johri
Dec 22 '18 at 7:20
add a comment |
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1
$begingroup$
Your proof is correct. Was that your question?
$endgroup$
– Jonas
Dec 22 '18 at 7:08
$begingroup$
I am always suspicious if there is a mistake :)
$endgroup$
– Dreamer123
Dec 22 '18 at 7:12
1
$begingroup$
In future, use the
proof-verificationtag$endgroup$
– Shubham Johri
Dec 22 '18 at 7:20