As part of an exercise I needed to decompose the Jordan form of a complex matrix into nilpotent and diagonalizable parts, and also explain why they commute. The decomposition is trivial since strictly triangular matrices are nilpotent, but I don't see - why do these matrices commute?
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.
43292
Let $a_1,\dots ,a_n \in \mathbb{R}^n$ with $a_i = (a_{i1}, \dots , a_{in})$.  Show that when $\mid a_{jj}\mid \gt \sum_{i \neq j} \mid a_{ij} \mid \; \forall \; j\in \{1,\dots ,n\}$, then $a_1,\dots ,a_n$ are linearly independent.  I want to show that the matrix \begin{matrix} a_{11} &amp; a_{21} &amp; \cdots &amp; a_{n1} \\ a_{12} &amp; a_{22} &amp; \cdots &amp; a_{n2} \\ \vdots &amp; &amp; \ddots &amp; \\ a_{1n} &amp; a_{2n} &amp; \cdots &amp; a_{nn} \end{matrix} can be transformed into an upper triangular matrix with all entries in the main diagonal $\neq 0$.  For the cases $n=1$ and $n=2$ this is easy. However, I fail to show this for the general case; the problem being that the tranformations mess up the coefficients such that I can't use the assumption on the $a_{jj}$'s anymore.  I'm pretty sure that this has already been solved somewhere on here, but I don't know how to search for it.
20168
I am currently reading two linear algebra books. One is Hoffman/Kunze's and the other one is Friedberg/Insel/Spence's.  They define Jordan canonical form of linear operator in different ways.  In Hoffman's book, Jordan forms are obtained by the primary decomposition and cyclic decomposition. In this case, 1's can appear some sub-diagonal entries. For example,  $M = \begin{array}{cc} 2 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\ 1 &amp; 2 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 3 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 3 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 3 \\ \end{array}$  But in Friedberg's (And almost any other linear algebra books for engineers..), 1's can appear some super-diagonal entries. For example,  $M' = \begin{array}{cc} 2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 2 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 3 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 3 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 3 \\ \end{array}$  Can someone explain why they define differently and relation of each other?
50716
Is it true for complex matrices $X,Y$ that $$ (XY)^*=X^* Y^*? $$ where $^*$ refers to complex conjugation.  How can we prove this if so?  Thanks!  Note: I am referring to complex conjugation, not the hermitan transpose. The answer below refers to hermitan tranpose.
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!
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
6236
Let A = $\begin{bmatrix}1 &amp; 1 &amp; 4\\0 &amp; 3 &amp; -4\\0&amp;0&amp;-1\end{bmatrix}$. Is the matrix A diagonalizable?  If so find a matrix P that diagonalizes A. Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors? Determine the eigendecomposition $A = PΛP^{-1}$. A = $\begin{bmatrix}3 &amp; 1 &amp; 0\\1 &amp; 2 &amp;1\\0&amp;1&amp;3\end{bmatrix}$. Is the matrix A diagonalizable?  If so find a matrix P that orthogonally diagonalizes A. Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors? (Note that A is real symmetric so that you do not have to compute the inverse of P).   My solution attempts:   det(λI - A) = $\begin{vmatrix}λ - 1 &amp; -1 &amp; -4\\0 &amp; λ - 3 &amp;4\\0&amp;0&amp;λ + 1\end{vmatrix}$ = (λ -1) (λ -3) (λ + 1) = 0   $λ_{1}$ = 1, $λ_{2}$ = 3, $λ_{3}$ = -1  ($λ_{1}I - A)x_{1}$ = 0  $\begin{bmatrix}0 &amp; -1 &amp; -4\\0 &amp; -2 &amp;4\\0&amp;0&amp;2\end{bmatrix}$ $\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}$ = $\begin{bmatrix}0\\0 \\0\end{bmatrix}$   $\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}$ = $x_{11} \begin{bmatrix}1\\0 \\0\end{bmatrix}$  It follows that $\begin{bmatrix}1\\0 \\0\end{bmatrix}$ is a basis for the eigenspace of A corresponding to $λ_{1}$ = 1.  In a similar way, I found other two basis vectors as $\begin{bmatrix}1/2\\1 \\0\end{bmatrix}$ and $\begin{bmatrix}-2/5\\1 \\1\end{bmatrix}$, and I got $P = \begin{bmatrix}1 &amp; 1/2 &amp; -2/5\\0 &amp; 1 &amp; 1\\0&amp;0&amp;1\end{bmatrix}$. Then my $P^{-1} = \begin{bmatrix}1 &amp; -1/2 &amp; 9/10\\0 &amp; 1 &amp; -1\\0&amp;0&amp;1\end{bmatrix}$. And I said A's left-multiplication by $P^{-1}$ and right-multiplication by $P (P^{-1}AP)$ gives the diagonal matrix Λ that is similar to A. I found Λ as $\begin{bmatrix}1 &amp; 0 &amp; 0\\0 &amp; 3 &amp; 0\\0&amp;0&amp;21/5\end{bmatrix}$. Then I said that if we multiply both sides of the equation $P^{-1}AP = Λ$ by $P$ from the left and by $P^{-1}$ from the right we get $PP^{-1}APP^{-1} = PΛP^{-1} = A$ and that this is the eigendecomposition of A. Here is my first question: Do you think I understood and did correctly what the asker wanted me to do by saying "Determine the eigendecomposition $A = PΛP^{-1}$."?   I started in the same manner and I got three eigenvectors as  $\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}$ = $x_{12} \begin{bmatrix}0\\1 \\0\end{bmatrix}$,  $\begin{bmatrix}x_{21}\\x_{22}\\x_{23}\end{bmatrix}$ = $x_{21} \begin{bmatrix}1\\1 \\1\end{bmatrix}$ and  $\begin{bmatrix}x_{31}\\x_{32}\\x_{33}\end{bmatrix}$ = $x_{33} \begin{bmatrix}1\\-2 \\1\end{bmatrix}$.   I said A is not diagonalizable since the eigenvectors are not linearly independent.  Here is my second question: What the hell does the sentence "Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors?" mean?
75026
Let $A$ be a complex $3\times 3$ matrix such that $A^3=-I$, then we need to find out which of the following statements are correct?   $A$ has three distinct eigenvalues; $A$ is diagonalizable over $\mathbb{C}$; $A$ is triangulizable over $\mathbb{C}$; $A$ is non singular.   Wll, from minimal polynomial approach I have deduced that all are correct as minpoly has to be $x^3+1$ am I right?
4880
If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$.    I do not understand anything more than the following.   Elementary row operations. Linear dependence. Row reduced forms and their relations with the original matrix.   If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?  P.S.: Please avoid using the transpose and/or inverse of a matrix.
58133
Firstly, I know this question might seem slightly vague, but I can't find any examples or specific questions to specify it.   My question is in regards to finding the Jordan Canonical Form of a matrix, more specifically when you have two Jordan chain of length greater than $1$ for a particular eigenvalue. After you've found the eigenvectors, are these eigenvectors always linearly independent? Also, if they are, is it possible to find eigenvectors which form Jordan chains which are 'wrong' because they aren't linear independent?
109840
Let us consider the $2n \times 2n$ block matrix $$\mathbf{X} = \pmatrix{\mathbf{P} &amp; \mathbf{Q}\\ \mathbf{R} &amp; \mathbf{S} }$$ where $\mathbf{P} , \mathbf{Q} , \mathbf{R} , \mathbf{S}$ are square  matrices of order $n$ which commute pairwise. Show that $$\det \mathbf{X} = \det (\mathbf{PS} - \mathbf{QR}).$$  The problem in this case I am having is when none of the matrices $ \mathbf{P} , \mathbf{Q} , \mathbf{R} , \mathbf{S} $ are invertible. Any help will be greatly appreciated.
104931
Problem: If $S$ is a skew-symmetric matrix, show that $(I+S)(I-S)^{-1}$ is orthogonal.  This appeared on a list of standard questions asked of Princeton graduate students. It has been a while since I've studied linear algebra, and frankly I cannot even see why $(I-S)$ must be invertible.
5337
I wish to prove the following:  if $T:X\to X$ is a nilpotent (and the     linear space $X$ is finite-dimensional), then there exists a basis     of $X$ such that the matrix representation of $T$ is upper     triangular with zero diagonal elements.  I was trying to prove $\{x_1,...,x_{k+1},v, Tv, ..., T^{n-k-2}v\}$ is a basis for $X$ where $x_i$ are basis for $N(T)$ and $v,..., T^{n-k-2}v$ are basis for $R(T)$ where $T^{n-k}=0$ and $n-k$ is minimal. However the matrix representation w.r.t this basis is diagonal. Is there an easy way to prove this ?  I know that for all $T:X\rightarrow X$, there exists a basis that the matrix representation is  upper triangular. How is this useful?
103141
Find all $2\times 2$ matrices that commute with  $$\left( \begin{array}{cc} 2 &amp; 3 \\ 1 &amp; 4 \end{array} \right)$$  My progress:  I know that a square matrix commutes with itself, the identity matrix of that order, the null matrix of that order and any scalar matrix of that order.  The answer has been given as:   $$\left( \begin{array}{cc} m &amp; 3n \\ n &amp; m+2n \end{array} \right)$$  I don't understand how they're getting that form. Can someone please explain?
91137
I'm struggling to understand the process of how to decompose a unitary transform into two-level unitary matrices. I've been trying to understand the process as detailed in , but I don't get where those 2x2 matrices in the individual two-level matrices are coming from, or what P actually is. If anyone could walk me through a numerical example of decomposing for example, a 4x4 matrix into two-level unitary matrices, that'd be greatly appreciated!
107921
Let , $S$ be the set of $3\times 3$ real matrices $A$ with $$AA^T=\left(\begin{matrix}1&amp;0&amp;0\\0&amp;0&amp;0\\0&amp;0&amp;0\end{matrix}\right).$$Then the set $S$ contains :   (A) a nilpotent matrix.  (B) a matrix of rank $1$.  (C) a matrix of rank $2$.  (D) a non-zero skew-symmetric matrix.  Attempt :  Clearly , $\left(\begin{matrix}1&amp;0&amp;0\\0&amp;0&amp;0\\0&amp;0&amp;0\end{matrix}\right)\in S$. So, (B) is TRUE.   Again $rank(AA^T)=rank(A)$. As , $rank(AA^T)=1$ , so $S$ does not contain a matrix of rank $2$. So (C) is FALSE.  Again we know the rank of a non-zero skew-symmetric matrix can never be $1$. So , $S$ does not contain a non-zero skew-symmetric matrix. So (D) is FALSE.  But I am unable to understand the option (A) whether it is correct or NOT ? Any hint ?
43402
Two $n\times n$ matrices $A, B$ are said to be simultaneously diagonalizable if there is a nonsingular matrix $S$ such that both $S^{-1}AS$ and $S^{-1}BS$ are diagonal matrices.      a) Show that simultaneously diagonalizable matrices commute: $AB = BA$.      b) Prove that the converse is valid, provided that one of the matrices has no multiple eigenvalues. Is every pair of commuting matrices simultaneously diagonalizable?   My attempt:  a) Let $M=S^{-1}AS$ and $P=S^{-1}BS$, then it follows that $A= S^{-1}MS$ and $B=S^{-1}PS$. Thus $AB =S^{-1}MSS^{-1}PS = S^{-1}MPS = S^{-1}PMS$ (is it wrong that I switched MP to PM?) $=S^{-1}PSS^{-1}MS=BA$.   b) How can I do this?