As I was studying linear algebra from various books I came to know about Cayley -Hamilton theorem. It states that: An $n×n$ matrix satisfies it's own characteristic equation. I see in various books that they have given a big proof for this. But this seems quite obvious to me. Because characteristic equation of an $n×n$ matrix A is given by: $det(xI_n-A)=0$, where $I_n$ is the $n×n$ identity matrix. If we put $x=A$ in the lhs of the characteristic equation we get: $det(A.I_n-A)=det(A-A)=det(O_n)=0$, where $O_n$ is the zero matrix of order n.  Is there any mistake in my idea? If not then why are they giving so much bigger proofs.
27203
Cayley Hamilton Theorem states that if $A$ is an $n \times n$ matrix over the field $F$ then $p(A) = 0$.  We note that $p(\lambda) = \det(\lambda I - A)$. Hence, why can't we just substitute $\lambda$ with $A$ and directly prove Cayley-Hamilton Theorem by saying that $p(A) = p(\lambda) = \det(\lambda I - A) = \det(AI - A) = 0$?
85546
I am having trouble proving that a matrix A and its transpose $$A^T$$ have the same eigenvalues. I tried to use the formula $\det(I\lambda - A)$ but am getting nowhere. Is anyone able to help?
123549
Let $A$ be an $n \times n$ complex matrix. Prove that there exist a diagonalizable matrix $D$ and a nilpotent matrix $N$ such that  a. A = D + N  b. DN = ND  and show that these matrices are uniquely determined.   I think I've solved the part a but don't have an idea to continue. Here is what I've tried: Let $D = \begin{bmatrix} a_{11} &amp; 0 \\ * &amp; a_{nn} \\ \end{bmatrix}$ and $N = \begin{bmatrix} 0 &amp; * \\ 0 &amp; 0 \end{bmatrix}$. Since D is a lower triangular matrix, its determinant is equal to product of its diagonal entries and hence its characteristic polynomial is $(x-a_{11})...(x-a_{nn}).$ Then D is diagonalizable. Similarly, characteristic polynomial of $N$ is $x^n$ then by Cayley-Hamilton $N^n = 0$. Update: My assumption for D to be diagonalizable was wrong, eigenvalues need not to be distinct. Update after answers: Thank you all for your help. It was just a question asked in the end of the chapter "Canonical Forms". So I just know Jordan form, Rational form, etc. I don't know anything about the Lie Algebra, Semi simple matrices, Representation theory, Perfect field mentioned in the answers. Honestly, answers didn't help me to understand the solution but they seem useful so maybe they help other people. Thanks.
1470
This question aims to create an &quot;&quot;  of numerous questions that ask about determinants of specific matrices (I may have missed a few):              The general question of this type is  Let $A$ be a square matrix of rank$~1$, let $I$ the identity matrix of the same size, and $\lambda$ a scalar. What is the determinant of $A+\lambda I$?  A clearly very closely related question is  What is the characteristic polynomial of a matrix $A$ of rank$~1$?
63340
For a given $n \times n$-matrix $A$, and $J\subseteq\{1,...,n\}$ let us denote by $A[J]$ its principal minor formed by the columns and rows with indices from $J$.  If the characteristic polynomial of $A$ is $x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, then why $$a_k=(-1)^{n-k}\sum_{|J|=n-k}A[J],$$ that is, why is each coefficient the sum of the appropriately sized principal minors of $A$?
12353
I am aware of some theorem that says that if $M$ is non singular, $\det(M) \neq 0$, then: $$\mathrm{card}(\mathbb{Z}^n/M \mathbb{Z}^n)= |\det(M)|.$$ How does one prove this? Figured if I put in the context of the rational canonical form, this would help, but cant piece it together.  Thanks.  This result is mentioned in the first answer of this question:  .
120572
Suppose we have $n\times n$ matrix $A$ having only positive elements and satisfying the property $a_{ij}=1/a_{ji}$ (a matrix satisfying this property is called a reciprocal matrix).  If its largest eigenvalue $λ_{max}$ is equal to $n$, then the matrix $A$ satisfies the property (consistency property) $a_{ij}a_{jk}=a_{ik}$ where $i,j,k=1,2,...,n$.    I already found an example of $5\times 5$ reciprocal matrix  to show this condition:  \begin{bmatrix}1&amp;1/2&amp;1&amp;1&amp;1/4\\2&amp;1&amp;2&amp;2&amp;1/2\\1&amp;1/2&amp;1&amp;1&amp;1/4\\1&amp;1/2&amp;1&amp;1&amp;1/4\\4&amp;2&amp;4&amp;4&amp;1\end{bmatrix}  It has simple characteristic equation $λ^5-5λ^4=0$, so the eigenvalues of matrix $A$ are $λ=0$ and $λ=5$ and it's shown that $λ_{max}=n=5$.  Is there anyone can help me to give another positive reciprocal matrix example  that has simple characteristic equation and integer number as its eigenvalues ($4\times 4, 5\times 5$ or $6\times 6$ matrix)?
99442
Let $A$ be a $4\times 4$ matrix over $\mathbb C$ such that $\operatorname{rank} A=2$ and $A^3=A^2\neq 0$  Suppose $A$ is not diagonalizable.  Then which one is true   $A^2=A\neq 0$ $\exists$ a vector $v$ such that $Av\neq 0$ but $A^2v= 0$ the characteristic polynomial of $A$ is $x^4-x^3$   My try:  If $A^2=A$ then minimal polynomial of $A$ has two distinct roots and hence $A$ is diagonalisable hence false. Since $A^3=A^2\neq 0$ so the characteristic roots of $A$ are $0,0,1$ Since rank of $A$ is $2$ $1$ must appear twice so the characteristic polynomial of $A=x^4-(0+0+1+1)x^3+(0.0.1+0.1.1+1.1.0+1.0.1)x^2+(0.0+0.1+1.1+...)x+(0.0.1.1)$ hence 3 is false  2 I am not sure Please find the error in my solution and don't refer it to some previous solution
8559
Hi guys I am working with this and I am trying to prove to myself that n by n matrices of the type zero on the diagonal and 1 everywhere else are invertible.  I ran some cases and looked at the determinant and came to the conclusion that we can easily find the determinant by using the following $\det(A)=(-1)^{n+1}(n-1)$. To prove this I do induction  n=2 we have the $A=\begin{bmatrix} 0 &amp; 1\\  1 &amp; 0 \end{bmatrix}$ $\det(A)=-1$ and my formula gives me the same thing (-1)(2-1)=-1  Now assume if for $n \times n$ and $\det(A)=(-1)^{n+1}(n-1)$  Now to show for  a matrix B of size $n+1 \times n+1$. I am not sure I was thinking to take the determinant of the $n \times n$ minors but I am maybe someone can help me. Also is there an easier way to see this is invertible other than the determinant? I am curious.
4726
I know that if A and B are square nxn matrices, then AB and BA have the same characteristic polynomial and thus the same eigenvalues (and same algebraïc multiplicity).   I'm wondering though if this can be generalized: if A is a nxm matrix and B a mxn matrix, then AB is a nxn matrix and BA a mxm matrix.  So my question is: will the eigenvalues of AB and BA, that differ from zero, have the same algebraïc multiplicity?
53474
Write down all the possible Jordan normal forms for matrices with characteristic polynomial $(x-a)^5$. In each case, calculate the minimal polynomial and the geometric multiplicity of the eigenvalue $a$. Verify that this information determines the Jordan normal form.  I found this question in a textbook that I'm using for a test I have tomorrow. I think that I need to use the method for finding the Jordan normal form of a matrix but I can't see how to apply it and I don't have much intuition about the answer...  I'm guessing that there are 5 possibilities since the minimal polynomial can be any factor of the characteristic, but I don't know how to prove this.  Some help would be great for my test tomorrow! Thanks
112968
How to prove the  determinant of $$A=(a_{ij})_{n\times n}=b_i^2-b_j^2$$ where $b_1,b_2,...,b_n$ are some distinct real numbers , $n\geq 3$  and  $i,j=1,2,...,n$  is zero?  Here,  $A$ is skew symmetric and I know odd order skew symmetric matrix has determinant zero. How about the even case? or Is there any other method to show $\det(A)=0$ ?
56520
I know how to solve "simple" recurrence relations. For instance, say you have:  $$c_0 = 20$$ $$c_1 = 30$$ $$c_n = 3 c_{n-1} - 2 c_{n-2}$$  We can write the characteristic equation as:  $$3x^{n-1} - 2x^{n-2} = x^n$$  Solving this with $n=2$, we get $x = 1$ or $x = 2$. This lets us write the relation $c_n = \alpha_1 1^n + \alpha_2 2^n$, and we can solve for $\alpha_1$ and $\alpha_2$ with the initial states $c_0$ and $c_1$.  However, this depends on the fact that $3x^{n-1} - 2x^{n-2} = x^n$ has two roots.  Now, I'm stuck on another problem where the characteristic equation has fewer roots than terms.  Say I have this recurrence relation instead:  $$a_0 = 0$$ $$a_1 = 2$$ $$a_2 = −1$$ $$a_n = 9a_{n-1} - 15a_{n-2} - 25a_{n-3}$$  The characteristic equation would be:  $$9x^{n-1} - 15x^{n-2} - 25x^{n-3} = x^n$$  However, solving with $n=3$, we only get two roots: $x=-1$ or $x=5$. There are not enough roots to write a relation in the form of $a_n = \alpha_1 r_1^n + \alpha_2r_2^n + \alpha_3r_3^n$. How do I proceed?
15554
It's a simple exercise to show that two similar matrices has the same eigenvalues and eigenvectors (my favorite way is noting that they represent the same linear transformation in different bases).   However, to show that two matrices has the same characteristic polynomial it does not suffice to show that they have the same eigenvalues and eigenvectors - one needs to say something smart about the algebraic multiplicities of the eigenvalues. Moreover, we might be working over a field which is not algebraically closed and hence simply "don't have" all the eigenvalues. This can be overcome, of course, by working in the algebraic closure of the field, but it complicates the explanation.  I'm looking for a proof that is simple and stand-alone as much as possible (the goal is writing an expository article about the subject, so clarity is the most important thing, not efficiency).
41104
I have to show that all $n \times n$ nilpotent matrices of order n are similar. My initial approach was to show that for all nilpotent matrices their minimun characteristic polynomial is of the form: $$\lambda^n$$  Is this sufficient?  Can someone show me a formal approach to this problem?  Thanks!
83594
Let $A$ be an $n\times n$ matrix. Then its characteristic polynomial is  $$\phi_A(x) = \det(xI - A) = x^n - (\operatorname{tr} A)x^{n-1} + \cdots + (-1)^{n-1}(\operatorname{tr}(\operatorname{adj}A))x + (-1)^n\det A,$$  where $\operatorname{adj}A$ denotes the  of $A$.     My question: Why are the coefficients of $\phi_A$ those shown above?   I can see why $(-1)^n\det A$ is the constant term, since $$\det(A) = \phi_A(0) = (0-\mu_1)\cdots(0-\mu_n) = (-1)^n\mu_1\cdots\mu_n,$$  where $\mu_i$ are the roots ($i=1,\dots,n$).  I also see why the coefficient of $x^{n-1}$ is minus the trace, since by Laplace expansion (along the first row):  \begin{align*}     &amp;\det(xI-A)\\[10pt]    &amp;= \begin{vmatrix}          x-a_{11} &amp; -a_{12} &amp; \cdots &amp; -a_{1n}\\          -a_{21} &amp; x-a_{22} &amp; \cdots &amp; -a_{2n}\\          \vdots &amp; \vdots &amp; \ddots &amp; \vdots\\          -a_{n1} &amp; -a_{n2} &amp; \cdots &amp; x-a_{nn}     \end{vmatrix}\\[10pt]     &amp;= (x-a_{11})\begin{vmatrix}          x-a_{22} &amp; \cdots &amp; -a_{2n}\\          \vdots &amp; \ddots &amp; \vdots\\          -a_{n2} &amp; \cdots &amp; x-a_{nn}     \end{vmatrix} + \underbrace{a_{12} \begin{vmatrix}          -a_{21} &amp; \cdots &amp; -a_{2n}\\          \vdots &amp; \ddots &amp; \vdots\\          -a_{n1} &amp; \cdots &amp; x-a_{nn}     \end{vmatrix} + \cdots}_{\text{at most polynomial of degree $n-2$}}\\[10pt]    &amp;=(x-a_{11})\bigg((x-a_{22})\begin{vmatrix}         x-a_{33} &amp; \cdots &amp; -a_{3n}\\        \vdots &amp; \ddots &amp; \vdots\\        -a_{n3} &amp; \cdots &amp; x-a_{nn}     \end{vmatrix}     +\underbrace{\underbrace{a_{23}\begin{vmatrix}         -a_{32} &amp; \cdots &amp; -a_{3n}\\        \vdots &amp; \ddots &amp; \vdots\\        -a_{n2} &amp; \cdots &amp; x-a_{nn}     \end{vmatrix}+\cdots}_{\text{at most polynomial of degree $n-3$}}      \bigg) + \cdots}_{\text{at most polynomial of degree $n-2$}}\\     &amp;\kern10pt\vdots\\[10pt]     &amp;=(x-a_{11})(x-a_{22})\cdots(x-a_{nn}) + \underbrace{\cdots}_{\text{at most polynomial of degree $n-2$}}\\[10pt]     &amp;= x^n - (a_{11}+a_{22}+\cdots + a_{nn})x^{n-1} + \cdots\\[10pt]     &amp;= x^n - (\operatorname{tr}A)x^{n-1} + \cdots \end{align*} (Let me know if there is a more elegant way to obtain this).   But what I cannot seem to prove is that the coefficient of $x$ is $ (-1)^{n-1}\operatorname{tr}(\operatorname{adj}A)$. Attempting to work with the Laplace expansion (as I did to obtain the trace above) is leading me nowhere.
115545
I'm having trouble following a proof of this generalization of the Cayley-Hamilton theorem:     Suppose that $M$ is an $A$-module generated by $n$ elements, and that $\varphi \in \operatorname{Hom}_A(M,M)$; let $I$ be an ideal of $A$ such that $\varphi(M) \subset IM$. Then there is a relation of the form $$(**)\quad \varphi^n + a_1 \varphi^{n-1} + \cdots + a_{n-1}\varphi + a_n = 0,$$ with $a_i \in I^i$ for $1 \leqslant i \leqslant n$ (where both sides are considered as endomorphisms of $M$).   This is theorem 2.1 in Matsumura's Commutative Ring Theory. The proof there goes as follows:     Let $M = A\omega_1 + \dots + A\omega_n$; by the assumption $\varphi(M) \subset IM$ there exist $a_{ij} \in I$ such that $\varphi(\omega_i) = \sum_{j=1}^n a_{ij}\omega_j$. This can be rewritten $$\sum_{j=1}^n(\varphi\delta_{ij} - a_{ij})\omega_j = 0 \quad (\text{for }\  1\leqslant i \leqslant n),$$ where $\delta_{ij}$ is the Kronecker symbol. The coefficients of this system of linear equations can be viewed as a square matrix $(\varphi\delta_{ij} - a_{ij})$ of elements of $A'[\varphi]$, the commutative subring of the endomorphism ring $E$ of $M$ generated by the image $A'$ of $A$ together with $\varphi$; let $b_{ij}$ denote its $(i,j)$th cofactor, and $d$ its determinant. By multiplying the above equation through by $b_{ik}$ and summing over $i$, we get $d\omega_k = 0$ for $1 \leqslant k \leqslant n$. Hence $d\cdot M = 0$, so that $d = 0$ as an element of $E$. Expanding the determinant $d$ gives a relation of the form $(**)$.   I would like to understand this particular proof, but I'm having trouble concluding $d\omega_k = 0$ for $1 \leqslant k \leqslant n$ from the multiplication of the above equation by $b_{ik}$ and summing over $i$. It's clear that the right hand side remains zero, but I'm having trouble expanding the left hand side into a useful form. I'm pretty sure I have to use the fact that $d = \sum_{i=1}^n b_{ik} (\varphi\delta_{ik} - a_{ik})$ by the cofactor expansion of the determinant along the $k$th column, but I can only seem conclude $d\omega_k - d\omega_k = 0$, which is not very enlightening.  Thanks in advance.