Find determinant and trace of product of non square matrices
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Let $B in M_{3,2}(mathbb{R})$,$C in M_{2,3}(mathbb{R})$ so that $BC=begin{pmatrix}2 & -2 & 3\
0 & 0 & 3\
0 & 0 & 3
end{pmatrix}$. Find $det(CB)$ and $Tr(CB)$.
My attempt : I tried to compute the powers of $BC$ (I actually hoped that $BC$ was idempotent) and I saw that
$(BC)^n=begin{pmatrix}2^n & -2^n & 3^n\
0 & 0 & 3^n\
0 & 0 & 3^n
end{pmatrix}$ which doesn't really help.
linear-algebra matrices
asked Dec 29 '18 at 13:46
Math GuyMath Guy
406
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add a comment |
$begingroup$
Let $B in M_{3,2}(mathbb{R})$,$C in M_{2,3}(mathbb{R})$ so that $BC=begin{pmatrix}2 & -2 & 3\
0 & 0 & 3\
0 & 0 & 3
end{pmatrix}$. Find $det(CB)$ and $Tr(CB)$.
My attempt : I tried to compute the powers of $BC$ (I actually hoped that $BC$ was idempotent) and I saw that
$(BC)^n=begin{pmatrix}2^n & -2^n & 3^n\
0 & 0 & 3^n\
0 & 0 & 3^n
end{pmatrix}$ which doesn't really help.
linear-algebra matrices
asked Dec 29 '18 at 13:46
Math GuyMath Guy
406
$endgroup$
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@PeterMelech $det(C)$ doesn't exist because are not squared matrix
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– Martín Vacas Vignolo
Dec 29 '18 at 14:08
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Yes sorry, I noticed already
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– Peter Melech
Dec 29 '18 at 14:08
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@Yanko the trace formula doesn't hold for any non square matrices even though in this case according to the answer key it does.
$endgroup$
– Math Guy
Dec 29 '18 at 14:13
$begingroup$
@MathGuy Ok I believe you, anyway I post an answer without using this.
$endgroup$
– Yanko
Dec 29 '18 at 14:18
add a comment |
$begingroup$
Let $B in M_{3,2}(mathbb{R})$,$C in M_{2,3}(mathbb{R})$ so that $BC=begin{pmatrix}2 & -2 & 3\
0 & 0 & 3\
0 & 0 & 3
end{pmatrix}$. Find $det(CB)$ and $Tr(CB)$.
My attempt : I tried to compute the powers of $BC$ (I actually hoped that $BC$ was idempotent) and I saw that
$(BC)^n=begin{pmatrix}2^n & -2^n & 3^n\
0 & 0 & 3^n\
0 & 0 & 3^n
end{pmatrix}$ which doesn't really help.
linear-algebra matrices
asked Dec 29 '18 at 13:46
Math GuyMath Guy
406
$endgroup$
Let $B in M_{3,2}(mathbb{R})$,$C in M_{2,3}(mathbb{R})$ so that $BC=begin{pmatrix}2 & -2 & 3\
0 & 0 & 3\
0 & 0 & 3
end{pmatrix}$. Find $det(CB)$ and $Tr(CB)$.
My attempt : I tried to compute the powers of $BC$ (I actually hoped that $BC$ was idempotent) and I saw that
$(BC)^n=begin{pmatrix}2^n & -2^n & 3^n\
0 & 0 & 3^n\
0 & 0 & 3^n
end{pmatrix}$ which doesn't really help.
linear-algebra matrices
linear-algebra matrices
asked Dec 29 '18 at 13:46
Math GuyMath Guy
406
asked Dec 29 '18 at 13:46
Math GuyMath Guy
406
edited Dec 29 '18 at 14:13
Math Guy
asked Dec 29 '18 at 13:46
Math GuyMath Guy
406
asked Dec 29 '18 at 13:46
Math GuyMath Guy
406
asked Dec 29 '18 at 13:46
Math GuyMath Guy
406
406
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@PeterMelech $det(C)$ doesn't exist because are not squared matrix
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– Martín Vacas Vignolo
Dec 29 '18 at 14:08
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Yes sorry, I noticed already
$endgroup$
– Peter Melech
Dec 29 '18 at 14:08
$begingroup$
@Yanko the trace formula doesn't hold for any non square matrices even though in this case according to the answer key it does.
$endgroup$
– Math Guy
Dec 29 '18 at 14:13
$begingroup$
@MathGuy Ok I believe you, anyway I post an answer without using this.
$endgroup$
– Yanko
Dec 29 '18 at 14:18
add a comment |
$begingroup$
@PeterMelech $det(C)$ doesn't exist because are not squared matrix
$endgroup$
– Martín Vacas Vignolo
Dec 29 '18 at 14:08
$begingroup$
Yes sorry, I noticed already
$endgroup$
– Peter Melech
Dec 29 '18 at 14:08
$begingroup$
@Yanko the trace formula doesn't hold for any non square matrices even though in this case according to the answer key it does.
$endgroup$
– Math Guy
Dec 29 '18 at 14:13
$begingroup$
@MathGuy Ok I believe you, anyway I post an answer without using this.
$endgroup$
– Yanko
Dec 29 '18 at 14:18
$begingroup$
@PeterMelech $det(C)$ doesn't exist because are not squared matrix
$endgroup$
– Martín Vacas Vignolo
Dec 29 '18 at 14:08
$begingroup$
@PeterMelech $det(C)$ doesn't exist because are not squared matrix
$endgroup$
– Martín Vacas Vignolo
Dec 29 '18 at 14:08
$begingroup$
Yes sorry, I noticed already
$endgroup$
– Peter Melech
Dec 29 '18 at 14:08
$begingroup$
Yes sorry, I noticed already
$endgroup$
– Peter Melech
Dec 29 '18 at 14:08
$begingroup$
@Yanko the trace formula doesn't hold for any non square matrices even though in this case according to the answer key it does.
$endgroup$
– Math Guy
Dec 29 '18 at 14:13
$begingroup$
@Yanko the trace formula doesn't hold for any non square matrices even though in this case according to the answer key it does.
$endgroup$
– Math Guy
Dec 29 '18 at 14:13
$begingroup$
@MathGuy Ok I believe you, anyway I post an answer without using this.
$endgroup$
– Yanko
Dec 29 '18 at 14:18
$begingroup$
@MathGuy Ok I believe you, anyway I post an answer without using this.
$endgroup$
– Yanko
Dec 29 '18 at 14:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You need to work with this:
The trace of a matrix is the sum of the eigenvalues and the determinant is the product.
It is not hard to see that the eigenvalues of $BC$ are $2,3,0$. Let $v_1,v_2$ be the eigenvectors correspond to $2,3$ (you will see why we can't use the eigenvector for $0$ soon).
Consider the matrix $CB$ and look at the vectors $Cv_1,Cv_2$. Then $$CB(Cv_i) = C(BCv_i)=lambda_i Cv_i$$ therefore since $Cv_inot = 0$ (otherwise $BCv_i=0$) we have that $2,3$ are eigenvalues of $CB$ which means that the determinant is $6$ and the trace is $5$.
answered Dec 29 '18 at 14:18
YankoYanko
6,5691728
$endgroup$
1
$begingroup$
Excellent solution,congratulations !
$endgroup$
– Math Guy
Dec 29 '18 at 14:22
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You need to work with this:
The trace of a matrix is the sum of the eigenvalues and the determinant is the product.
It is not hard to see that the eigenvalues of $BC$ are $2,3,0$. Let $v_1,v_2$ be the eigenvectors correspond to $2,3$ (you will see why we can't use the eigenvector for $0$ soon).
Consider the matrix $CB$ and look at the vectors $Cv_1,Cv_2$. Then $$CB(Cv_i) = C(BCv_i)=lambda_i Cv_i$$ therefore since $Cv_inot = 0$ (otherwise $BCv_i=0$) we have that $2,3$ are eigenvalues of $CB$ which means that the determinant is $6$ and the trace is $5$.
answered Dec 29 '18 at 14:18
YankoYanko
6,5691728
$endgroup$
1
$begingroup$
Excellent solution,congratulations !
$endgroup$
– Math Guy
Dec 29 '18 at 14:22
add a comment |
$begingroup$
You need to work with this:
The trace of a matrix is the sum of the eigenvalues and the determinant is the product.
It is not hard to see that the eigenvalues of $BC$ are $2,3,0$. Let $v_1,v_2$ be the eigenvectors correspond to $2,3$ (you will see why we can't use the eigenvector for $0$ soon).
Consider the matrix $CB$ and look at the vectors $Cv_1,Cv_2$. Then $$CB(Cv_i) = C(BCv_i)=lambda_i Cv_i$$ therefore since $Cv_inot = 0$ (otherwise $BCv_i=0$) we have that $2,3$ are eigenvalues of $CB$ which means that the determinant is $6$ and the trace is $5$.
answered Dec 29 '18 at 14:18
YankoYanko
6,5691728
$endgroup$
1
$begingroup$
Excellent solution,congratulations !
$endgroup$
– Math Guy
Dec 29 '18 at 14:22
add a comment |
$begingroup$
You need to work with this:
The trace of a matrix is the sum of the eigenvalues and the determinant is the product.
It is not hard to see that the eigenvalues of $BC$ are $2,3,0$. Let $v_1,v_2$ be the eigenvectors correspond to $2,3$ (you will see why we can't use the eigenvector for $0$ soon).
Consider the matrix $CB$ and look at the vectors $Cv_1,Cv_2$. Then $$CB(Cv_i) = C(BCv_i)=lambda_i Cv_i$$ therefore since $Cv_inot = 0$ (otherwise $BCv_i=0$) we have that $2,3$ are eigenvalues of $CB$ which means that the determinant is $6$ and the trace is $5$.
answered Dec 29 '18 at 14:18
YankoYanko
6,5691728
$endgroup$
You need to work with this:
The trace of a matrix is the sum of the eigenvalues and the determinant is the product.
It is not hard to see that the eigenvalues of $BC$ are $2,3,0$. Let $v_1,v_2$ be the eigenvectors correspond to $2,3$ (you will see why we can't use the eigenvector for $0$ soon).
Consider the matrix $CB$ and look at the vectors $Cv_1,Cv_2$. Then $$CB(Cv_i) = C(BCv_i)=lambda_i Cv_i$$ therefore since $Cv_inot = 0$ (otherwise $BCv_i=0$) we have that $2,3$ are eigenvalues of $CB$ which means that the determinant is $6$ and the trace is $5$.
answered Dec 29 '18 at 14:18
YankoYanko
6,5691728
answered Dec 29 '18 at 14:18
YankoYanko
6,5691728
answered Dec 29 '18 at 14:18
YankoYanko
6,5691728
answered Dec 29 '18 at 14:18
YankoYanko
6,5691728
6,5691728
1
$begingroup$
Excellent solution,congratulations !
$endgroup$
– Math Guy
Dec 29 '18 at 14:22
add a comment |
1
$begingroup$
Excellent solution,congratulations !
$endgroup$
– Math Guy
Dec 29 '18 at 14:22
1
1
$begingroup$
Excellent solution,congratulations !
$endgroup$
– Math Guy
Dec 29 '18 at 14:22
$begingroup$
Excellent solution,congratulations !
$endgroup$
– Math Guy
Dec 29 '18 at 14:22
add a comment |
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$begingroup$
@PeterMelech $det(C)$ doesn't exist because are not squared matrix
$endgroup$
– Martín Vacas Vignolo
Dec 29 '18 at 14:08
$begingroup$
Yes sorry, I noticed already
$endgroup$
– Peter Melech
Dec 29 '18 at 14:08
$begingroup$
@Yanko the trace formula doesn't hold for any non square matrices even though in this case according to the answer key it does.
$endgroup$
– Math Guy
Dec 29 '18 at 14:13
$begingroup$
@MathGuy Ok I believe you, anyway I post an answer without using this.
$endgroup$
– Yanko
Dec 29 '18 at 14:18