Find determinant and trace of product of non square matrices












2














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







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  • @PeterMelech $det(C)$ doesn't exist because are not squared matrix
    – Martín Vacas Vignolo
    1 hour ago










  • Yes sorry, I noticed already
    – Peter Melech
    1 hour ago










  • @Yanko the trace formula doesn't hold for any non square matrices even though in this case according to the answer key it does.
    – Math Guy
    1 hour ago












  • @MathGuy Ok I believe you, anyway I post an answer without using this.
    – Yanko
    1 hour ago
















2














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
























  • @PeterMelech $det(C)$ doesn't exist because are not squared matrix
    – Martín Vacas Vignolo
    1 hour ago










  • Yes sorry, I noticed already
    – Peter Melech
    1 hour ago










  • @Yanko the trace formula doesn't hold for any non square matrices even though in this case according to the answer key it does.
    – Math Guy
    1 hour ago












  • @MathGuy Ok I believe you, anyway I post an answer without using this.
    – Yanko
    1 hour ago














2












2








2







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















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 1 hour ago

























asked 2 hours ago









Math Guy

405




405












  • @PeterMelech $det(C)$ doesn't exist because are not squared matrix
    – Martín Vacas Vignolo
    1 hour ago










  • Yes sorry, I noticed already
    – Peter Melech
    1 hour ago










  • @Yanko the trace formula doesn't hold for any non square matrices even though in this case according to the answer key it does.
    – Math Guy
    1 hour ago












  • @MathGuy Ok I believe you, anyway I post an answer without using this.
    – Yanko
    1 hour ago


















  • @PeterMelech $det(C)$ doesn't exist because are not squared matrix
    – Martín Vacas Vignolo
    1 hour ago










  • Yes sorry, I noticed already
    – Peter Melech
    1 hour ago










  • @Yanko the trace formula doesn't hold for any non square matrices even though in this case according to the answer key it does.
    – Math Guy
    1 hour ago












  • @MathGuy Ok I believe you, anyway I post an answer without using this.
    – Yanko
    1 hour ago
















@PeterMelech $det(C)$ doesn't exist because are not squared matrix
– Martín Vacas Vignolo
1 hour ago




@PeterMelech $det(C)$ doesn't exist because are not squared matrix
– Martín Vacas Vignolo
1 hour ago












Yes sorry, I noticed already
– Peter Melech
1 hour ago




Yes sorry, I noticed already
– Peter Melech
1 hour ago












@Yanko the trace formula doesn't hold for any non square matrices even though in this case according to the answer key it does.
– Math Guy
1 hour ago






@Yanko the trace formula doesn't hold for any non square matrices even though in this case according to the answer key it does.
– Math Guy
1 hour ago














@MathGuy Ok I believe you, anyway I post an answer without using this.
– Yanko
1 hour ago




@MathGuy Ok I believe you, anyway I post an answer without using this.
– Yanko
1 hour ago










1 Answer
1






active

oldest

votes


















4














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

















  • 1




    Excellent solution,congratulations !
    – Math Guy
    1 hour ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4














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

















  • 1




    Excellent solution,congratulations !
    – Math Guy
    1 hour ago
















4














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

















  • 1




    Excellent solution,congratulations !
    – Math Guy
    1 hour ago














4












4








4






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












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 1 hour ago









Yanko

6,029724




6,029724








  • 1




    Excellent solution,congratulations !
    – Math Guy
    1 hour ago














  • 1




    Excellent solution,congratulations !
    – Math Guy
    1 hour ago








1




1




Excellent solution,congratulations !
– Math Guy
1 hour ago




Excellent solution,congratulations !
– Math Guy
1 hour ago


















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