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Kernel of polynomial of matrix
Linear Algebra: Minimum PolynomialCharacteristic Polynomial of a Linear MapPolynomial of matrixImage and Kernel of a MatrixKernel of a polynomialA positive definite matrix A can be express as a polynomial of $ A^2 $How does minimal polynomial divide characteristic polynomial of matrix?Kernel and Image of matricesFinding a matrix given its characteristic polynomialChange of basis of the kernel of a rectangular matrix
$begingroup$
I am given that $f(x)$, $g(x)$ and $h(x)$ are polynomials such that $f(x)=g(x)h(x)$.
I am then asked to show using induction on the degree of $f$ that $f(A)$ would be an $n times n$-matrix if $A$ is an $n times n$ matrix. I am told any constant term of $f(x), c,$ will be replaced by $cI$ in the matrix polynomial $f(A)$, where $I$ is the $n times n$ identity matrix - same size as A.
Wouldn’t this be trivial as if $I$ is the $n times n$ identity matrix, A would need to be the same size in order to add to I and thus any linear combination of powers of A would be the same size? I don’t know what induction has to do with this.
I am also asked to show that if $f(x)=g(x) h(x)$ and $f(A)=0$, then for any vector $v$ and any polynomial $b(A)$, then $b(A) h(A) v inker(g(A))$ and, similarly, $b(A) g(A) v inker(h(A))$. I don’t know how to start because I think that $b(A)h(A)v$ and $b(A)g(A)v$ are both vectors and we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.
Any hints or help would be greatly appreciated!
linear-algebra matrices polynomials
asked Mar 16 at 10:02
AndrewAndrew
357213
$endgroup$
add a comment |
$begingroup$
I am given that $f(x)$, $g(x)$ and $h(x)$ are polynomials such that $f(x)=g(x)h(x)$.
I am then asked to show using induction on the degree of $f$ that $f(A)$ would be an $n times n$-matrix if $A$ is an $n times n$ matrix. I am told any constant term of $f(x), c,$ will be replaced by $cI$ in the matrix polynomial $f(A)$, where $I$ is the $n times n$ identity matrix - same size as A.
Wouldn’t this be trivial as if $I$ is the $n times n$ identity matrix, A would need to be the same size in order to add to I and thus any linear combination of powers of A would be the same size? I don’t know what induction has to do with this.
I am also asked to show that if $f(x)=g(x) h(x)$ and $f(A)=0$, then for any vector $v$ and any polynomial $b(A)$, then $b(A) h(A) v inker(g(A))$ and, similarly, $b(A) g(A) v inker(h(A))$. I don’t know how to start because I think that $b(A)h(A)v$ and $b(A)g(A)v$ are both vectors and we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.
Any hints or help would be greatly appreciated!
linear-algebra matrices polynomials
asked Mar 16 at 10:02
AndrewAndrew
357213
$endgroup$
add a comment |
$begingroup$
I am given that $f(x)$, $g(x)$ and $h(x)$ are polynomials such that $f(x)=g(x)h(x)$.
I am then asked to show using induction on the degree of $f$ that $f(A)$ would be an $n times n$-matrix if $A$ is an $n times n$ matrix. I am told any constant term of $f(x), c,$ will be replaced by $cI$ in the matrix polynomial $f(A)$, where $I$ is the $n times n$ identity matrix - same size as A.
Wouldn’t this be trivial as if $I$ is the $n times n$ identity matrix, A would need to be the same size in order to add to I and thus any linear combination of powers of A would be the same size? I don’t know what induction has to do with this.
I am also asked to show that if $f(x)=g(x) h(x)$ and $f(A)=0$, then for any vector $v$ and any polynomial $b(A)$, then $b(A) h(A) v inker(g(A))$ and, similarly, $b(A) g(A) v inker(h(A))$. I don’t know how to start because I think that $b(A)h(A)v$ and $b(A)g(A)v$ are both vectors and we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.
Any hints or help would be greatly appreciated!
linear-algebra matrices polynomials
asked Mar 16 at 10:02
AndrewAndrew
357213
$endgroup$
I am given that $f(x)$, $g(x)$ and $h(x)$ are polynomials such that $f(x)=g(x)h(x)$.
I am then asked to show using induction on the degree of $f$ that $f(A)$ would be an $n times n$-matrix if $A$ is an $n times n$ matrix. I am told any constant term of $f(x), c,$ will be replaced by $cI$ in the matrix polynomial $f(A)$, where $I$ is the $n times n$ identity matrix - same size as A.
Wouldn’t this be trivial as if $I$ is the $n times n$ identity matrix, A would need to be the same size in order to add to I and thus any linear combination of powers of A would be the same size? I don’t know what induction has to do with this.
I am also asked to show that if $f(x)=g(x) h(x)$ and $f(A)=0$, then for any vector $v$ and any polynomial $b(A)$, then $b(A) h(A) v inker(g(A))$ and, similarly, $b(A) g(A) v inker(h(A))$. I don’t know how to start because I think that $b(A)h(A)v$ and $b(A)g(A)v$ are both vectors and we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.
Any hints or help would be greatly appreciated!
linear-algebra matrices polynomials
linear-algebra matrices polynomials
asked Mar 16 at 10:02
AndrewAndrew
357213
asked Mar 16 at 10:02
AndrewAndrew
357213
edited Mar 16 at 10:32
Andrew
asked Mar 16 at 10:02
AndrewAndrew
357213
asked Mar 16 at 10:02
AndrewAndrew
357213
asked Mar 16 at 10:02
AndrewAndrew
357213
357213
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To show that $f(A)$ is an $ntimes n$-matrix, you could prove by induction that $A^k$ is an $ntimes n$-matrix for all $kinBbbN$, and then note that any linear combination of $ntimes n$-matrices is again an $ntimes n$-matrix. I agree that it's a rather trivial thing to prove though.
To show that $b(A)h(A)vinker g(A)$ for any polynomial $b$ and any vector $v$, it suffices to show that $h(A)vinker g(A)$. You claim that
...we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.
Indeed you can't take powers of vectors, but that's not what is being claimed.
You have just shown (by induction) that $g(A)$ is an $ntimes n$-matrix. So you can multiply it by an $n$-vector. Because $h(A)$ is also an $ntimes n$-matrix $h(A)v$ is an $n$-vector. So you can multiply $g(A)$ by $h(A)v$. To show that $h(A)vinker g(A)$ it suffices to show that the product is $0$.
answered Mar 16 at 10:23
ServaesServaes
28.5k34099
$endgroup$
$begingroup$
Thank you. I can’t use the null factor law, though, can I? That is if $f(A)=g(A)h(A)=0$ this does not imply that either $g(A)$ or $h(A)$ is equal to zero because two nonzero matrices can multiply to zero. Sorry that I’m the original question I forgot to add that $f(A)=0$ for this second part.
$endgroup$
– Andrew
Mar 16 at 10:31
$begingroup$
@Andrew I assumed that $f(A)=0$ for the second part as otherwise it doesn't make sense. And indeed, $g(A)h(A)=0$ does not imply that $g(A)=0$ or $h(A)=0$. But you do know that $g(A)cdot h(A)v=0$ for all $v$...
$endgroup$
– Servaes
Mar 16 at 10:33
$begingroup$
Of course! Thank you very much.
$endgroup$
– Andrew
Mar 16 at 10:38
add a comment |
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$begingroup$
To show that $f(A)$ is an $ntimes n$-matrix, you could prove by induction that $A^k$ is an $ntimes n$-matrix for all $kinBbbN$, and then note that any linear combination of $ntimes n$-matrices is again an $ntimes n$-matrix. I agree that it's a rather trivial thing to prove though.
To show that $b(A)h(A)vinker g(A)$ for any polynomial $b$ and any vector $v$, it suffices to show that $h(A)vinker g(A)$. You claim that
...we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.
Indeed you can't take powers of vectors, but that's not what is being claimed.
You have just shown (by induction) that $g(A)$ is an $ntimes n$-matrix. So you can multiply it by an $n$-vector. Because $h(A)$ is also an $ntimes n$-matrix $h(A)v$ is an $n$-vector. So you can multiply $g(A)$ by $h(A)v$. To show that $h(A)vinker g(A)$ it suffices to show that the product is $0$.
answered Mar 16 at 10:23
ServaesServaes
28.5k34099
$endgroup$
$begingroup$
Thank you. I can’t use the null factor law, though, can I? That is if $f(A)=g(A)h(A)=0$ this does not imply that either $g(A)$ or $h(A)$ is equal to zero because two nonzero matrices can multiply to zero. Sorry that I’m the original question I forgot to add that $f(A)=0$ for this second part.
$endgroup$
– Andrew
Mar 16 at 10:31
$begingroup$
@Andrew I assumed that $f(A)=0$ for the second part as otherwise it doesn't make sense. And indeed, $g(A)h(A)=0$ does not imply that $g(A)=0$ or $h(A)=0$. But you do know that $g(A)cdot h(A)v=0$ for all $v$...
$endgroup$
– Servaes
Mar 16 at 10:33
$begingroup$
Of course! Thank you very much.
$endgroup$
– Andrew
Mar 16 at 10:38
add a comment |
$begingroup$
To show that $f(A)$ is an $ntimes n$-matrix, you could prove by induction that $A^k$ is an $ntimes n$-matrix for all $kinBbbN$, and then note that any linear combination of $ntimes n$-matrices is again an $ntimes n$-matrix. I agree that it's a rather trivial thing to prove though.
To show that $b(A)h(A)vinker g(A)$ for any polynomial $b$ and any vector $v$, it suffices to show that $h(A)vinker g(A)$. You claim that
...we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.
Indeed you can't take powers of vectors, but that's not what is being claimed.
You have just shown (by induction) that $g(A)$ is an $ntimes n$-matrix. So you can multiply it by an $n$-vector. Because $h(A)$ is also an $ntimes n$-matrix $h(A)v$ is an $n$-vector. So you can multiply $g(A)$ by $h(A)v$. To show that $h(A)vinker g(A)$ it suffices to show that the product is $0$.
answered Mar 16 at 10:23
ServaesServaes
28.5k34099
$endgroup$
$begingroup$
Thank you. I can’t use the null factor law, though, can I? That is if $f(A)=g(A)h(A)=0$ this does not imply that either $g(A)$ or $h(A)$ is equal to zero because two nonzero matrices can multiply to zero. Sorry that I’m the original question I forgot to add that $f(A)=0$ for this second part.
$endgroup$
– Andrew
Mar 16 at 10:31
$begingroup$
@Andrew I assumed that $f(A)=0$ for the second part as otherwise it doesn't make sense. And indeed, $g(A)h(A)=0$ does not imply that $g(A)=0$ or $h(A)=0$. But you do know that $g(A)cdot h(A)v=0$ for all $v$...
$endgroup$
– Servaes
Mar 16 at 10:33
$begingroup$
Of course! Thank you very much.
$endgroup$
– Andrew
Mar 16 at 10:38
add a comment |
$begingroup$
To show that $f(A)$ is an $ntimes n$-matrix, you could prove by induction that $A^k$ is an $ntimes n$-matrix for all $kinBbbN$, and then note that any linear combination of $ntimes n$-matrices is again an $ntimes n$-matrix. I agree that it's a rather trivial thing to prove though.
To show that $b(A)h(A)vinker g(A)$ for any polynomial $b$ and any vector $v$, it suffices to show that $h(A)vinker g(A)$. You claim that
...we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.
Indeed you can't take powers of vectors, but that's not what is being claimed.
You have just shown (by induction) that $g(A)$ is an $ntimes n$-matrix. So you can multiply it by an $n$-vector. Because $h(A)$ is also an $ntimes n$-matrix $h(A)v$ is an $n$-vector. So you can multiply $g(A)$ by $h(A)v$. To show that $h(A)vinker g(A)$ it suffices to show that the product is $0$.
answered Mar 16 at 10:23
ServaesServaes
28.5k34099
$endgroup$
To show that $f(A)$ is an $ntimes n$-matrix, you could prove by induction that $A^k$ is an $ntimes n$-matrix for all $kinBbbN$, and then note that any linear combination of $ntimes n$-matrices is again an $ntimes n$-matrix. I agree that it's a rather trivial thing to prove though.
To show that $b(A)h(A)vinker g(A)$ for any polynomial $b$ and any vector $v$, it suffices to show that $h(A)vinker g(A)$. You claim that
...we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.
Indeed you can't take powers of vectors, but that's not what is being claimed.
You have just shown (by induction) that $g(A)$ is an $ntimes n$-matrix. So you can multiply it by an $n$-vector. Because $h(A)$ is also an $ntimes n$-matrix $h(A)v$ is an $n$-vector. So you can multiply $g(A)$ by $h(A)v$. To show that $h(A)vinker g(A)$ it suffices to show that the product is $0$.
answered Mar 16 at 10:23
ServaesServaes
28.5k34099
answered Mar 16 at 10:23
ServaesServaes
28.5k34099
answered Mar 16 at 10:23
ServaesServaes
28.5k34099
answered Mar 16 at 10:23
ServaesServaes
28.5k34099
28.5k34099
$begingroup$
Thank you. I can’t use the null factor law, though, can I? That is if $f(A)=g(A)h(A)=0$ this does not imply that either $g(A)$ or $h(A)$ is equal to zero because two nonzero matrices can multiply to zero. Sorry that I’m the original question I forgot to add that $f(A)=0$ for this second part.
$endgroup$
– Andrew
Mar 16 at 10:31
$begingroup$
@Andrew I assumed that $f(A)=0$ for the second part as otherwise it doesn't make sense. And indeed, $g(A)h(A)=0$ does not imply that $g(A)=0$ or $h(A)=0$. But you do know that $g(A)cdot h(A)v=0$ for all $v$...
$endgroup$
– Servaes
Mar 16 at 10:33
$begingroup$
Of course! Thank you very much.
$endgroup$
– Andrew
Mar 16 at 10:38
add a comment |
$begingroup$
Thank you. I can’t use the null factor law, though, can I? That is if $f(A)=g(A)h(A)=0$ this does not imply that either $g(A)$ or $h(A)$ is equal to zero because two nonzero matrices can multiply to zero. Sorry that I’m the original question I forgot to add that $f(A)=0$ for this second part.
$endgroup$
– Andrew
Mar 16 at 10:31
$begingroup$
@Andrew I assumed that $f(A)=0$ for the second part as otherwise it doesn't make sense. And indeed, $g(A)h(A)=0$ does not imply that $g(A)=0$ or $h(A)=0$. But you do know that $g(A)cdot h(A)v=0$ for all $v$...
$endgroup$
– Servaes
Mar 16 at 10:33
$begingroup$
Of course! Thank you very much.
$endgroup$
– Andrew
Mar 16 at 10:38
$begingroup$
Thank you. I can’t use the null factor law, though, can I? That is if $f(A)=g(A)h(A)=0$ this does not imply that either $g(A)$ or $h(A)$ is equal to zero because two nonzero matrices can multiply to zero. Sorry that I’m the original question I forgot to add that $f(A)=0$ for this second part.
$endgroup$
– Andrew
Mar 16 at 10:31
$begingroup$
Thank you. I can’t use the null factor law, though, can I? That is if $f(A)=g(A)h(A)=0$ this does not imply that either $g(A)$ or $h(A)$ is equal to zero because two nonzero matrices can multiply to zero. Sorry that I’m the original question I forgot to add that $f(A)=0$ for this second part.
$endgroup$
– Andrew
Mar 16 at 10:31
$begingroup$
@Andrew I assumed that $f(A)=0$ for the second part as otherwise it doesn't make sense. And indeed, $g(A)h(A)=0$ does not imply that $g(A)=0$ or $h(A)=0$. But you do know that $g(A)cdot h(A)v=0$ for all $v$...
$endgroup$
– Servaes
Mar 16 at 10:33
$begingroup$
@Andrew I assumed that $f(A)=0$ for the second part as otherwise it doesn't make sense. And indeed, $g(A)h(A)=0$ does not imply that $g(A)=0$ or $h(A)=0$. But you do know that $g(A)cdot h(A)v=0$ for all $v$...
$endgroup$
– Servaes
Mar 16 at 10:33
$begingroup$
Of course! Thank you very much.
$endgroup$
– Andrew
Mar 16 at 10:38
$begingroup$
Of course! Thank you very much.
$endgroup$
– Andrew
Mar 16 at 10:38
add a comment |
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