For example if i have a linear operator $\ T:\mathbb{R}^n\to \mathbb{R}^n$ and some basis $\ B=\{v_1,v_2,..v_n\}$.  Let $\ [T]_B$ be the representation matrix of $\ T $ with respect to the basis $\ B$. for example if I do $\ [T]_B*v_2 $ do I get the coordinates of $\ v_2$ in the standard basis?
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?
14820
I'm having a hard time truly understanding the meaning of $\dim\operatorname{Im} T + \dim\operatorname{Ker} T = \dim V$ where $V$ is the domain of a linear transformation $T:V\to W$. I've used this equation several times in many problems, and I've gone over the proof and I believe that I fully understand it, but I don't understand the intuitive reasoning behind it. I'd appreciate an intuitive explanation of it.  Just to be clear, I do understand the equation itself, I am able to use it, and I know how to prove it; my question is what is the meaning of this equation from a linear algebra perspective.
92364
I'm reading Sheldon Axler's "Linear Algebra Done Right", and I have a question about one of the examples he gives on page 130. Let $T$ be a linear operator on $F^2$ whose matrix (with respect to the standard basis) is $$\begin{bmatrix} 2 &amp; -3 \\ 3 &amp; 2\end{bmatrix}$$ I can see why this operator is not self-adjoint, but I can't see why it is normal. I have to find $T^*$ and show that $TT^* = T^*T$. How would you do this?
119288
This question has been doing my head in, as it limits us to not use eigenvalues, diagonalization or anything analogous -- just strictly linear functions.     Given a linear function $f : \mathbb R^3 \to \mathbb R^3$ that satisfies $p(x) = x^2 - x + 1,$ find the matrix $A$ of $f$ with respect to the standard basis of $\mathbb R^3$
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!
62196
I am trying to solve the following problem:   Let $A$ be an $n\times n$ matrix, and let $V$ denote the space of $n$-dimensional row vectors. What is the matrix of the linear operator ‘‘right multiplication by $A$’’ with respect to the standard basis of $V$?  I am not sure where to begin with this problem.
48940
We're given: $$ V = Span\left\{ \begin{bmatrix} 2 \\ 2 \\ 2 \\ 1 \end{bmatrix}, \ \begin{bmatrix} 2 \\ 1 \\ 1 \\ 0 \end{bmatrix}, \ \begin{bmatrix} 5 \\ 4 \\ 1 \\ 1 \end{bmatrix}  \right\} \ \ \   \mathrm{and} \ \ \ W = Span \left\{\begin{bmatrix}1\\-3\\2\\1 \end{bmatrix} \right\} $$  and we're asked to find $\mathrm{dim}(V\cap W^{\bot})$  Here's my approach. First, by inspecting the basis of $W$, I managed to construct a basis for $W^{\bot}$, which is the following: $$W^{\bot} = Span \left\{\begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 1 \\ 0 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}\right\}$$ Then, for each vector $\textbf{w}$ in the basis of $W^{\bot}$, I tried to see if the system $A\textbf{x}=\textbf{w}$ was compatible or not. In this case, I found out that the system was only compatible with two of the vectors in $W^{\bot}$, thus indicating me that $\mathrm{dim}(V \cap W^{\bot} ) = 2$ (which is correct).  What I do not get, however, is  that since the dimension of the intersection is $2$, why aren't two of the vectors in the basis of $V$ orthogonal to the vector which spans $W$ (that was my initial approach, i.e try to see which vectors of the basis of $V$ are orthogonal to the vector that spans $W$).  Also, I was wondering if there was a simpler/quicker way of doing this.
96858
So I'm trying to prove the classic dimension theorem for linear transformations. Here is what I have so far:  Given a linear transformation $T:V\rightarrow W$  Let the basis of $\mathrm{Im}T$ be $\{u_1,...,u_m\}$ where $\forall u\in V:T(u)$.  Let the basis of $\ker T$ be $\{v_1,..,v_k\}$ where $\forall v:T(v)=0$.  Lets prove $B=\{u_1,...,u_m,v_1,..,v_k\}$ is a basis of $V$:  (1) $B$ is linearly independent: Assume (falsely) that $B$ isn't linearly independent. Since $\{u_1,...,u_m\}$ and $\{v_1,..,v_k\}$ are both the basis of $\ker T$ and $\mathrm{Im}T$ repectively, they must each be linearly independent.If this is the case, $\exists b\in V:b\in \{v_1,..,v_k\}\cap \{v_1,..,v_k\}$ however this contradicts $\mathrm{Im}T\cap \ker T=\emptyset$ so $B$ must be linearly independent.  (2)$\mathrm{span}(B)=V$: Assume (falsely) that $b_2\in V| b_2\notin \mathrm{span}(B)$. If this is the case, then $b_2\notin \{\forall u\in V:T(u)\}\cap\{\forall v:T(v)=0\}$. This is impossible because $b_2\in V$ so $\mathrm{span}(B)=V$  Since (1) and (2) are true, according to basis identities, $B$ is a basis of $V$. This means that $\dim V=|B|=\dim\ker T+\dim\mathrm{Im}T$ QED
58056
Let $T: V \rightarrow V$ be a linear operator.  I need to demonstrate that if all nonzero vectors of $V$ are eigenvectors of $T$, then there is one specific $\lambda \in K$ such that $T(v) = \lambda v$, for all $v \in V$.  I understand that, if all nonzero vectors of $V$ are eigenvectors of $T$, then $T$ must be a scaling transformation. It just stretch or shrinks vectors, but doesn't change their  directions.  So, the statement says that if it happens, then, there is a single $\lambda$ such that $T(v) = \lambda v$. In other words, if there is such transformation, then it scales all vectors by the same scalar $\lambda$.  Applying the transformation to our standard basis vectors, we have: $$ T(e_1) = \lambda_1 e_1 \\ T(e_2) = \lambda_2 e_2 \\ \vdots \\ T(e_n) = \lambda_n e_n $$  I understand I need to prove that $\lambda_1 = \lambda_2 = \dots = \lambda_n$, but I can't see how!  EDIT  $$ v = c_1e_1 + c_2e_2 + \dots + c_ne_n \\ T(v) = \mu v = \lambda_1c_1e_1 + \lambda_2c_2e_2 + \dots + \lambda_nc_ne_n \\ $$  Since what's multiplying $v$ coordinates is $\lambda_i$, then all of them must be $\mu$. I'm not sure how to 'mathematize' this. Is this idea correct?  EDIT 2 Extending the left hand side of EDIT 1, we have: $$ \mu v = \lambda_1c_1e_1 + \dots + \lambda_nc_ne_n \\ \mu(c_1e_1 + \dots + c_ne_n) = \lambda_1c_1e_1 + \dots + \lambda_nc_ne_n \\ \mu c_1e_1 + \dots + \mu c_ne_n = \lambda_1c_1e_1 + \dots + \lambda_nc_ne_n \\ $$  And since $e_i$ are linearly independent, $\mu = \lambda_1 = \lambda_2 = \dots = \lambda_n$. Is this proof correct?
15674
Let $V$ be a real n dimensional vector space &amp; let $T:V→V$ be a linear transformation satisfying $T^2(v)=-v$ for all $v\in V$. Then how can we show that $n$ is even?  I am completely stuck on it. Can anybody help me please?
63468
This is a very elementary question but I can't find the answer in my book at the moment.  If I have, for example, two vectors $v_1$ and $v_2$ in $\mathbb R^5$ and knowing that they are linear independent , how can I extend this two vectors to a basis of $\mathbb R^5$. I  know that I just have to add $v_i$ for $3\leq i \leq 5$ such that they are linear independent but how can I do that? Is there an easy algorithm?
115181
Question:  Suppose $V$ and $W$ are finite dimensional vector spaces over $\mathbb{C}$. $A$ is a linear transformation on $V$,  $B$ is a linear transformation on $W$. Then there exists a non-zero linear map $C:V\to W$ s.t. $CA=BC$ iff $A$ and $B$ have a same eigenvalue.  ===========This is incorrect==========  Clearly, if $CA=BC$, suppose $Ax=\lambda x$, then $B(Cx)=CAx=C(\lambda x)=\lambda (Cx)$, so $A$ and $B$ have same eigenvalue $\lambda$. On the other hand, if $A$ and $B$ have a same eigenvalue $\lambda$, suppose $Ax=\lambda x, By=\lambda y$. Define $C:V\to W$ s.t. $Cx=y$, then $BCx=By=\lambda y=C\lambda x=CAx$. But I don't kow how to make $BC=CA$ for all $x\in V$.   ======================================
47790
Let $(e_1,e_2)$ be the standard basis of $\mathbb{R}^2$. Why can the element $e_1\otimes e_2 + e_2\otimes e_1$ in $ \mathbb{R}^2\otimes_{\mathbb{R}} \mathbb{R}^2 $ not be written as a simple tensor $v\otimes w$ for $v,w\in \mathbb{R}^2$?  If we suppose $e_1\otimes e_2 + e_2\otimes e_1=v\otimes w$, then can we conclude that $e_1=\lambda e_2$ for some $\lambda \in \mathbb{R}$? This would give a contradiction since $e_1$ and $e_2$ are linearly independent. However, I am not sure if the argument is valid.
28584
I am currently learning group theory and I learnt that the fundamental representation and the anti-fundamental representation of $\text{SL}(2,\mathbb{C})$, $2 \times 2$ matrix with determinant of $1$, are not equivalent.  This means that no similarity transformation can map one of them to the other.  My professor gave an explanation (on the 2nd last paragraph on page 75 of the following document ) but I don't see how the difference in the signs in the exponent imply that the representations are inequivalent.  Can anyone please explain the explanation of my professor, or perhaps give another explanation?
39552
Use $\cal L$ to denote a linear transformation on some vector space. We know any matrix $\bf{A}$ can be viewed as a linear transformation by defining $\cal L:= \cal L(\bf{v})= Av$ where $\bf{v}$ is a vector. I am curious is any linear transformation can be represented by a matrix? If so, why? If not, can anyone help provide a counterexample?  In particular, projection of vectors onto a vector $\bf{b}$ is a linear transformation defined by ${\cal L(\bf{v})}=({\mathbf{v}} \cdot \frac{{\mathbf{b}}}{{|{\mathbf{b}}|}})\frac{{\mathbf{b}}}{{|{\mathbf{b}}|}}$. Can this linear transformation be represented by a matrix?  Thank you!
39873
I'm sorry if this is a silly question. I'm new to the notion of bases and all the examples I've dealt with before have involved sets of vectors containing real numbers. This has led me to assume that bases, by definition, are made up of a number of $n$-tuples.  However, now I've been thinking about a basis for all $n\times n$ matrices and I keep coming back to the idea that the simplest basis would be $n^2$ matrices, each with a single $1$ in a unique position.  Is this a valid basis? Or should I be trying to get column vectors on their own somehow?
120290
I'm trying to find a basis for the kernel for the following mapping:  Considering the linear transformation T: $M_{33} \rightarrow M_{33}  $ defined by $T(A) = .5(A + A^T)$. I know that this is basically asking for the basis under the condition that $T(A)=0$ which means that $A+A^T=0$ so $A^T = -A$.  I found that matrices that fit this condition are Skew Symmetric Matrices.  However, I'm not sure how to find the basis for the kernel of these matrices.