I know that it is possible to read the basis for $\operatorname{Null}(A)$ and $\operatorname{Null}(A^T)$ by simply looking at the reduced row-echelon form (RREF) of the matrix $A$. I have only an approximate idea of how to do it.  For example, if we $$A = \begin{bmatrix} 1 &amp; 2  &amp; 0  &amp; -1 \\  0 &amp; 0 &amp;1  &amp;2 \\  0 &amp;0  &amp;0  &amp;0  \end{bmatrix},$$ then one of the basis vectors for $\operatorname{Null}(A)$ will be $\begin{bmatrix}-2 &amp; 1 &amp; 0 &amp;0\end{bmatrix}^T$, and the other one will be $\begin{bmatrix}1 &amp; 0 &amp; -2 &amp;1 \end{bmatrix}^T$. So, basically we're "going around" over eight digits and alternating signs. But I noticed that this method isn't perfect. Does someone have a better idea of how to read bases for these two spaces from the RREF of $A$?
61627
The following text is written in my textbook and I don't really understand it:     If $A = (a_{ij}) \in$ Mat$(m x N, F)$ is a matrix in reduced row echelon form with $r$ nonzero rows and pivots in the columns numbered $j_1 &lt; ... &lt; j_r$, then the kernel ker$(A)$ is generated by the $n-r$ elements $ w_k = e_k - \sum\limits_{1 \le i \le r, j_i \le k} a_{ik}e_{j_i}$ for $k \in {1, \cdots , n} \ {j_1, \cdots, j_r}$, where $e_1, \cdots, e_n$ are the standard generators of $F^n$.    There are no computational examples and the notation is a bit overwhelming so I don't know how I can use it. Can somebody give an example of this in practice?
56883
Came across a problem in my textbook and I am having trouble with it, the book provides an answer but no workings. Looking to get an explanation of what is going on  help I prepare for an exam. If we let $T : \mathbb R^2 \to \mathbb R^2$ be a linear transformation. Consdider two bases $V$ and $W$ of the vector space $\mathbb R^2$, where $V=\{v_1,v_2\}$ with $ v_1 =  \begin{bmatrix} -2 \\1   \end{bmatrix}$ and  $ v_2 =  \begin{bmatrix} 1 \\1   \end{bmatrix}$ and where $W = \{w_1,w_2\}$ with  $ w_1 =  \begin{bmatrix} -1 \\4   \end{bmatrix}$ and  $ w_2 =  \begin{bmatrix} 0 \\1   \end{bmatrix}$. Suppose that  $ A =  \begin{bmatrix} 1&amp;2   \\-1 &amp;4    \end{bmatrix}$ is the matrix of $T$ under basis $W$. I need to find the matrix $B$ the matrix of $T$ under basis $V$.  The solution is:   $ B = P^{-1}AP$ = $   \begin{bmatrix} -30&amp;22   \\-48 &amp; 35    \end{bmatrix} $  But I don't know why.
108527
Let $A$ be an $n \times n$ matrix with real enteries. Define $\langle x,y\rangle_A = \langle Ax,Ay\rangle, \quad x,y \in \mathbb R^n$ , where $\langle,\rangle$ is a standard inner product on $\mathbb R^n$. Then  $\langle x,y\rangle_A$ is an inner product iff  1)  $\ker A = \{0\}$.  2) $\operatorname{rank} A= n$.  3) all eigenvalues of $A$ are positive.  4) all eigen value of $A$ are non negative.  Clearly if $\ker A \neq \{0\}$, then there exist $x \neq 0$ such that $Ax =0$, then $\langle Ax, Ax\rangle_A = 0$, then $\langle ,\rangle_A$ is not inner product, so $\ker A = \{0\}$, thus rank of $A$ is $n$.  I am unable to check the 3) and 4) option. Please help to check the other option. Thank you
63028
Suppose we have vector spaces $\mathbb W$ and $\mathbb V$.  A given basis for $\mathbb W$ is $\{(1, 1, 0, -1), (0, 1, 3, 1)\}$, and a given basis for $\mathbb V$ is $\{(1, 2, 2, -2), (0, 1, 2, -1)\}$.  How to find a basis for $\mathbb W\cap\mathbb V$?
99495
So if I have an $n\times n$ matrix $A$ that is diagonalizable then how do I determine $\mbox{tr}(A)$ and  $\det(A)$ as functions of the eigenvalues of the matrix?   For  $\det(A)$, I know the formula for eigenvalues is $\det(A−tI)$ which would be  $$(-1)^n(t-\lambda_1)\dots (t-\lambda_n)$$ where $\lambda_i$ are the eigenvalues of $A$.   For $\mbox{tr}(A)$ could I also do something similar to that? I'm just not sure how this question works.
59281
Suppose we have a $n \times m$ rectangular matrix $A$ with row space $\text{row}(A)$ and column space $\text{col}(A)$.   What are the row spaces and the column spaces of $A^T A$ and $A A^T$?
87131
"Find rank of matrix $\begin{bmatrix}                     0&amp;0&amp;-3\\                      9&amp;3&amp;5\\                     3&amp;1&amp;1\\                     \end{bmatrix}$ using echelon form?" "I am getting an answer equal to 2."  "When I reduce this I get $\begin{bmatrix}                           1&amp;1/3&amp;0\\                           0&amp;0&amp;1\\                            0&amp;0&amp;0\\ \end{bmatrix}$ so two pivots so rank must me 2 right ? However the book I am using says rank =1."
33526
I am searching for a short coordinate-free  proof of $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$ for linear operators  $A$, $B$ between finite dimensional vector spaces of the same dimension.    The usual proof is to represent the operators as matrices and then use matrix multiplication. I want a coordinate-free proof. That is, one that does not  make reference to an explicit matrix representation of the operator. I define trace as the sum of the eigenvalues of an operator.  Ideally, the proof the should be shorter and require fewer preliminary lemmas than the one given in .  I would be especially interested in a proof that generalizes to the trace class of operators on a Hilbert space.
91537
As everyone know that we can use a matrix $A$ to represent an operator $T$.  The adjoint of a matrix $A$ is denoted as $A^*$, which takes complex conjugate of $A$ and then transpose.      My problem is, what is the relationship between $A^*$ and $A^{-1}$?    I mean as we know $Ax = b$, $x = A^{-1}b$. (If not emphasize on the dimension of $x$ and $b$.
109630
We have been given matrix: $$A =          \begin{pmatrix}         a&amp; b&amp; c &amp;d \\         b &amp;−a&amp; d&amp; −c\\         c&amp; −d &amp;−a&amp; b \\         d &amp;c&amp; −b&amp; −a\\         \end{pmatrix}     $$ ...and have been asked to calculate $\det(A)$ using $AA^T$.  We see that:  $$AA^T=\begin{pmatrix}         a^2+ b^2+ c^2+ d^2&amp; 0&amp; 0&amp;0\\         0 &amp;a^2+ b^2+ c^2+ d^2&amp; 0&amp; 0\\         0&amp; 0 &amp;a^2+ b^2+ c^2+ d^2&amp; b \\        0 &amp;0&amp; 0&amp; a^2+ b^2+ c^2+ d^2\\         \end{pmatrix}     $$   So, $\det(AA^T)= (a^2+ b^2+ c^2+ d^2)^4$  Now which should I choose: $(a^2+ b^2+ c^2+ d^2)^2 $ or $ -(a^2+ b^2+ c^2+ d^2)^2$ ?  Please explain me which one and why.
122168
Given the basis $\beta = \{ (1, 1, 0),\ (1, 0, -1),\ (2, 1, 0)\}$ and the matrix:  $$ A =         \begin{bmatrix}         1 &amp; 1 &amp; 0 \\         1 &amp; 1 &amp; 0 \\         0 &amp; 0 &amp; 1 \\         \end{bmatrix} $$  with respect to  $\beta$.  How to change this matrix with respect to standard basis?
39803
This is the fourth part of a four-part problem in Charles W. Curtis's book entitled Linear Algebra, An Introductory Approach (p. 216).  I've succeeded in proving the first three parts, but the most interesting part of the problem eludes me.  Part (a) requires the reader to prove that $\operatorname{Tr}{(AB)} = \operatorname{Tr}{(BA)}$, which I was able to show by writing out each side of the equation using sigma notation.  Part (b) asks the reader to use part (a) to show that similar matrices have the same trace.  If $A$ and $B$ are similar, then  $\operatorname{Tr}{(A)} = \operatorname{Tr}{(S^{-1}BS)}$  $= \operatorname{Tr}(BSS^{-1})$  $= \operatorname{Tr}(B)$,  which completes part (b).  Part (c) asks the reader to show that the vector subspace of matrices with trace equal to zero have dimension $n^2 - 1$.  Curtis provides the hint that the map from $M_n(F)$ to $F$ is a linear transformation.  From this, I used the theorem that $\dim T(V) + \dim n(T) = \dim V$ to obtain the dimension of the null space.  Part (d), however, I'm stuck on.  It asks the reader to show that subspace described in part (c) is generated by matrices of the form $AB - BA$, where $A$ and $B$ are arbitrary $n \times n$ matrices.  I tried to form a basis for the subspace, but wasn't really sure what it would look like since an $n \times n$ matrix has $n^2$ entries in it, but the basis would need $n^2 - 1$ matrixes.  I also tried to think of a linear transformation whose image would have the form of $AB - BA$, but this also didn't help me.  I'm kind of stuck...  Many thanks in advance!
105770
Let us look the the matrix  $\left(\begin{array}{ccccc} a &amp; b &amp; b &amp; b &amp; b\\ b &amp; a &amp; b &amp; b &amp; b\\ b &amp; b &amp; a &amp; b &amp; b\\ b &amp; b &amp; b &amp; a &amp; b\\ b &amp; b &amp; b &amp; b &amp; a \end{array}\right)$  It has one value, $a$, on the main diagonal, and another value, $b$ everywhere else. Let us assume that we are over a ring and that $a$ is invertible.  I wish to find the determinant of every $n\times n$ matrix of this form ($a$ on the diagonal, $b$ everywhere else). Using row and column operations I have managed to transform the matrix to upper-triangular form and found formula for specific cases. Generalizing it I got to the following formula:  $\det(A) = a\left(a-b\right)^{n-2}\left(a+\left(n-2\right)b-\frac{\left(n-1\right)b^{2}}{a}\right)$  I think I can prove it with row-operations in the general case with a little patience.  However, I'm wondering if there is a "smart" way of getting to this formula that I'm missing and if there is a nicer representation of it. Also, what can be said when $a$ is not invertible? (esp. the case where we are over a field and $a=0$).
71019
Consider the vectorspace of all real $m \times n$ vectors and define an    inner product $\langle A,B\rangle = \operatorname{tr}(B^T   A)$.    "tr" stands for "trace"        which is the sum of the diagonal entries of a matrix.   How do you prove that $\operatorname{tr}(B^T   A)$ is indeed a inner product?  Kind regards
120486
The question on the homework wants a $2\times2$ matrix $R$, but $R$ not equal to positive or negative identity, that when you raise to seventh power you get the identity.  $R^7=\operatorname{Id}$.  I thought it would have to be some ones and negative ones,  it no luck, then tried some ones and some $.5$.  Just using an online matrix calculator to do brute force.  We just started talking about eigenvalues with visual example, so assuming wouldn't need that since we have not talked about how to find them yet.  Any hints or ideas to get me back on track?
92627
I have a matrix:  $$ A=\begin{bmatrix}1 &amp; a \\ 0 &amp; 1\end{bmatrix} $$  which satisfies: $A\left( e_{1}\right)=e_{1},A\left( e_{2}\right)=ae_{1}+e_{2}, a\neq 0$  I need compute all powers $A^{n}, n\in \mathbb{Z}$ and find their matrices.  $$ A^n= \begin{bmatrix}1 &amp; na \\ 0 &amp; 1\end{bmatrix} $$  Is it true?  Is there a basis of $V$ (if $V$ is a 2-dimensional real vector space with basis $\left\{e_{1},e_{2}\right\}$) so that the matrix of $A$ is diagonal?
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?