Find determinant and trace of product of non square matrices












2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    @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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    @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














2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 14:13







Math Guy

















asked Dec 29 '18 at 13:46









Math GuyMath Guy

576




576












  • $begingroup$
    @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










  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















4












$begingroup$

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 $5$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Excellent solution,congratulations !
    $endgroup$
    – Math Guy
    Dec 29 '18 at 14:22











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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 $5$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Excellent solution,congratulations !
    $endgroup$
    – Math Guy
    Dec 29 '18 at 14:22
















4












$begingroup$

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 $5$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Excellent solution,congratulations !
    $endgroup$
    – Math Guy
    Dec 29 '18 at 14:22














4












4








4





$begingroup$

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 $5$.






share|cite|improve this answer









$endgroup$



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 $5$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 14:18









YankoYanko

7,9651830




7,9651830








  • 1




    $begingroup$
    Excellent solution,congratulations !
    $endgroup$
    – Math Guy
    Dec 29 '18 at 14:22














  • 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


















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