What is the dimension of the space of $n\times n$ matrices with real entries which are such that the sum of the entries in the first row and the sum of the diagonal entries are both zero?? I tried but I get $n-1$ I don't know it is right or wrong??
121509
What is the dimension of the space of all $n \times n$ matrices with real entries   which are such that the sum of the entries in the first row and the sum of   the diagonal entries are both zero?   I tried by finding number of independent entries. The number of independent entries on the diagonal is $n-1$. The number of upper triangular independent entries is $\frac{n(n-1)}{2}-1(n-1)$, and the number of lowertriangular independent entries is $\frac{n(n-1)}{2}$. Now adding them will give dimension. Am I right?
35361
Given $A=$ $\left\{ {(1,2 + i,i),(1,3 + i,3 - i),(i,3i,4 + i)} \right\}$  Let $SP_CA$ be the linear space spanned by A over $C$  Let $SP_RA$ be the linear space spanned by A over $R$  what is the dimension of each subspace ?  I did row operations on A and saw that A is linearly dependet.  and that $dim(SP_CA)=2$  But over $R$. I cannot do row operations with the scalar $i$ since its not belong to $R$ so what is $dim(SP_RA)$ ?
38895
The following is an interesting problem from Linear Algebra 2nd Ed - Hoffman &amp; Kunze (3.5 Q17).   Let $W$ be the subspace spanned by the commutators of $M_{n\times n}\left(F\right)$: $$C=\left[A, B\right] = AB-BA$$  Prove that $W$ is exactly the subspace of matrices with zero trace.  Assuming this is true, one can construct $n^2 - 1$ linearly independent matrices, in particular $$[e_{i,n}, e_{n,i}]\ \text{for $1\le i\le n-1$}$$ $$[e_{i,n}, e_{j,n}]\ \text{for $i\neq j$}$$ where $e_{i,j}$ are the standard basis with $0$ entry everywhere except row $i$ column $j$ which span the space of traceless matrices.  However, I have trouble showing (or rather, believing, since this fact seems to be given) that the set of commutators form a subspace. In particular, I am having difficulty showing that the set is closed under addition. Can anyone shed some light?
35038
I ran into this problem recently.  If $A$ is a $n\times n$ matrix with $1$s in the diagonal and all off diagonal entries have absolute value less than $1$, is $A$ invertible?  It it definitely true for a $2\times 2$ matrix and seems to be true for larger matrices, just by trying to compute the determinant, however, my expression seems to get messy, and I was wondering if there was a easier method to solve this. If not, direction on how to argue from definition of determinant would be appreciated. Although I can see that an argument might come from there, I am having problems with that too.  Thank you for any help, and any help is much appreciated.
85090
Let $M_n(\mathbb{C})$ denote the vector space over $\mathbb{C}$ of all $n\times n$ complex matrices. Prove that if $M$ is a complex $n\times n$ matrix then $C(M)=\{A\in M_n(\mathbb{C}) \mid AM=MA\}$ is a subspace of dimension at least $n$.   My Try:  I proved that $C(M)$ is a subspace. But how can I show that it is of dimension at least $n$. No idea how to do it. I found similar questions posted in MSE but could not find a clear answer. So, please do not mark this as duplicate.  Can somebody please help me how to find this?   EDIT: Non of the given answers were clear to me. I would appreciate if somebody check my try below:  If $J$ is a Jordan Canonical form of $A$, then they are similar. Similar matrices have same rank. $J$ has dimension at least $n$. So does $A$. Am I correct?
101311
Let $V$ be the vector space made by the $n \times n$ square matrices.  a) Prove that $S=\{A\in V|A^t=A\}$ is a subspace of $V$  b) Prove that $T=\{A\in V|A^t=-A\}$ is a subspace of $V$  c) Prove that $V=S\oplus T$    SOLUTION  a) Let $O$ be the zero matrix, then $O \in S$ because $O^t=O$.  Then, let $A, B \in S$; and $r, k \in \mathbb R$. Hence  $$rA+kB=r(A^t)+k(B^t)=rA^t+kB^t$$ then $rA+kB \in S \implies$ $S$ is a subspace of $V$  b) Let $O$ be the zero matrix, then $O \in T$ because $O^t=O$.  Then, let $A, B \in T$; and $r, k \in \mathbb R$. Hence  $$rA+kB=r(-A^t)+k(-B^t)=-rA^t-kB^t=-(rA^t+kB^t)$$ then $rA+kB \in T \implies$ $T$ is a subspace of $V$  c) I have no idea how to do it.  I HOPE YOU CAN HELP ME VERYFING THE a), b) parts and with the part c).
123394
Let $V$ be an $n$-dimensional vector space over $\mathbb R$. Prove that $V$ has a subspace of dimension $r$ for each $0 \le r \le n$.  Is this as simple as saying that $V$ has a basis of $n$ elements, and then take $r$ elements from this basis and this will generate an $r$-dimensional subspace? Or am I missing something? It seems too simple.  Thanks
91498
So I was working on a specific problem related to Hermitian matrices. If we let $H_n$ denote the set of n x n Hermitian matrices. We're told that $H_n$ is a real vector space under matrix addition and scalar multiplication by a real number. I don't understand why though, because what would prevent you from adding two imaginary numbers?   Then we're told to write a basis for $H_2$. Now for a Hermitian matrix, it's equal to its conjugate transpose, so a basis would be:  $$\left(\begin{array}{ccc} 0 &amp; 1 \\ 1 &amp; 0\\ \end{array}\right) , \left(\begin{array}{ccc} 0 &amp; 0 \\ 0 &amp; 1\\ \end{array}\right), \left(\begin{array}{ccc} 1 &amp; 0 \\ 0 &amp; 0\\ \end{array}\right), \left(\begin{array}{ccc} 0 &amp; i \\ -i &amp; 0\\ \end{array}\right)$$  Which should be the same as the basis for a symmetric matrix, correct?   From there though, we have $sH_n$ represent skew-Hermitian matrices. I believe the basis for $sH_n$ is:  Edit: updated my basis here.   $$\left(\begin{array}{ccc} 0 &amp; -1 \\ 1 &amp; 0\\ \end{array}\right), \left(\begin{array}{ccc} 0 &amp; i \\ i &amp; 0\\ \end{array}\right)$$  because the diagonals would have to be zero right, to make the conjugate transpose of A be equal to -A?  From there though, they ask if $sH_n$ is a real vector space, and I'm not sure what to answer. It's real if we only use reals?  I'm asked the same thing for $U_n$, where $U_n$ represents n x n unitary matrices. I'm not even sure how to come up with a basis for this set??  Edit: Technically, the question says "Is $U_n$ a real vector space? Write a basis if possible."  I suppose I couldn't write a basis because $U_n$ is not a real vector space then. How would I show that though?   Thanks in advance for your help.
72416
Let $E$ be an $n$-dimensional vector space and $F_1,\ldots, F_k$ be linear subspaces of $E$ such that $$\sum_{i=1}^k \dim F_i &gt; n(k-1)$$ Prove $\displaystyle \cap_{i=1}^k F_i \neq \{0\}$.  I find this problem quite puzzling. The condition $\sum_{i=1}^k \dim F_i &gt; n(k-1)$ says that $\sum_{i=1}^k \dim F_i$ is close to its best upper bound so the $F_i$ have relatively big dimensions. I haven't made any progress on this one. I feel there's some trick involved.
102062
I had a question in a recent assignment that asked if a $3\times 3$ matrix could have a null space equal to its column space... clearly no, by the rank+nullity theorem... but I have a hard time wrapping my head around the concept of such a matrix, no matter what size, even existing. How could this be possible, and does anybody have an example of such a $m\times n$ matrix?
49426
I am looking to do the following:     Construct an inner product space $X$ (with inner product $\langle \cdot, \cdot \rangle$ and a proper, closed subspace $Y$ of $X$ such that $Y^\perp = \{0\}$, ie $\langle x, y \rangle = 0 \ \forall y \in Y \iff x = 0$.   I see that we need $X$ to be infinite dimensional, hence isomorphic to $\Bbb R^n$ with the standard dot product, and then clearly not possible, as a proper (closed) subspace has a smaller dimension.  If someone could give me a hint as to how to start this construction, then I'd be most grateful, as I'm fairly stuck beyond this! As always, please make sure that the hint is reasonably minor, ie doesn't give away too much - I still want to learn from this question, not just be told the answer!  The question is on a course in linear analysis.
122053
Find a basis for the vector space of symmetric matrices with an order of $n \times n$  This is my thought: by definition of symmetry, $a_{i,j}=a_{j,i}$. Therefore, the basis should consist ${n^2-n} \over 2$ matrices to determine each symmetric pair. In addition, it should also consist $n$ matrices to determine each term in the diagonal.    Therefore, the dimension of the vector space is ${n^2+n} \over 2$.   It's not hard to write down the above mathematically (in case it's true).    Two questions:     Am I right? Is that the desired basis? Is there a more efficent alternative to reprsent the basis?     Thanks!
24766
Let $A\neq I$ be a $5\times5$ matrix with real entries such that the sum of the entries in each row of $A$ is $1$. Then the sum of all the entries in $A^3$ is  1)$\space 3$ $\qquad $2)$\space 15$ $\qquad$ 3)$\space 5$ $\qquad$ 4)$\space 125$  Solution:  The answer is $5$ if $A = I$, but for $A \neq I$,  I found that vectors $\begin{pmatrix} 1\\ 1\\ 1\\ 1\\ 1\\ \end{pmatrix}$ and  $\begin{pmatrix} -1\\ -1\\ -1\\ -1\\ -1\\ \end{pmatrix}$  are eigen vectors for the matrix $A$ corresponding to eigen value $1$. From here, I dont know how to proceed, please help me solve the question..and if there are mistakes in my argument, please tell me..
108421
Let $V$ be the set of all symmetric $3 \times 3$ matrices. (Recall that $V$ is a subspace of $M_{3\times 3}.$) Find a basis of $V$ , and show that it is a basis.  I'm not sure what the question means by "basis of $V$." Any help would be appreciated.
120871
Can someone please check my proof and my definitions.  Let $A \in \mathbb{R}^{n \times m}$ be my matrix.  The left null space of $A$ is written as,  $\mathcal{N}(A^\top) = \{x \in \mathbb{R}^n| A^\top x = 0\}$  The orthogonal complement of the column space $\mathcal{C}(A)$ is written as,  $\mathcal{C}(A)^\perp = \{x \in \mathbb{R}^n | x^\top y = 0, \forall y \in \mathcal{C}(A)\}$     We want to show that $\mathcal{N}(A^\top)  = \mathcal{C}(A)^\perp $   First, we show, $\mathcal{N}(A^\top) \subseteq \mathcal{C}(A)^\perp$  Let $x \in \mathcal{N}(A^\top)$, then $A^\top x = 0 \implies x^\top A = 0^\top \implies x^\top Av= 0^\top v, \forall v \in \mathcal{C}(A) \implies x^\top y = 0 , y = Av$, $\implies x \in C(A)^\perp$.  Next, we show, $\mathcal{N}(A^\top) \supseteq \mathcal{C}(A)^\perp$  Let $x \in C(A)^\perp$, then $x^\top y = 0$, forall $y \in C(A)$. But $y = Av, \forall v \in \mathbb{R}^n$. Hence, $x^\top y = x^\top Av = v^\top A^\top x.$ For all $v \neq 0, A^\top x = 0$, hence $x \in \mathcal{N}(A^\top)$.    I'm pretty confident about the first proof. But the second proof is a bit more rough. Can someone please check for me.
39048
Looking for some help with the following question. I have begun to understand the concept of subspaces, but not sure how to deal with transverse matrices.  So I need to determine which of the subsets of $M_n(\Bbb R)$ are actually subspaces:  1) $ U = \{A \in M_n\Bbb (R) : A^t = A\}$, 2) $ U = \{A \in M_n\Bbb (R) : A^t = -A\}$, 3) $ U = \{A \in M_n\Bbb (R) : A^t \neq A\}$.
48440
Suppose the matrices $A$ and $B$ commute. Do there exists sequences $A_n$ and $B_n$ of matrices such that   $A_n \rightarrow A$, $B_n \rightarrow B$.  Each $A_n$ is diagonalizable  and the same for each $B_n$.  For every $n$, $A_n$ commutes with $B_n$.    Moreover, it would be nice if the following property was additionally satisfied: if $A,B$ are real, then $A_n,B_n$ can be chosen to be real as well.