Let $N(A^\top)$ be the left null space of $A$  $N(A^\top) = \{x| A^\top x = 0\}$  Let $C(A)^\perp$ be the orthogonal complement of the column space of $A$.  $C(A)^\perp = \{y| y^\top x = 0 \text{ for all } x \text{ in the column space of A}\}$  I want to show that they are equal.  In particular, I want to show that  $C(A)^\perp \subseteq N(A^\top)$    Proof attempt:  Let $y \in C(A)^\perp$, then $y^\top x = 0$.   Since $x$ is in the column space of $A$, there exists some vector $v$ such that $x = Av$, hence $y^\top (Av) = 0$.   This is the same as writing $(y^\top A) v = 0$ or $(A^\top y)^\top v = 0$.    I want to conclude that this implies $A^\top y = 0$ and hence $y \in N(A^\top)$. But this doesn't necessarily have to be the case.  First, $v$ could be zero.   Second, even if $v$ is not zero, you can have the case $(A^\top y)^\top = [-1, 1]$ and $v = [1, -1]^\top$, so $(A^\top y)^\top v = 0$ even though $A^\top y \neq 0$  Can someone please help?
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.
45157
Let $V$ be a vector space over  $\mathbb{C}$. If $V$ is an inner product space, then $V$  is normed (where the norm is defined as $\|x\|=\sqrt{(x,x)}\,\,$). Now if $V$ is normed, does it follow that $V$ is an inner product space ? I suspect no. I would like to see an example.  Thank you.    After reading my question again, I think it needs some clarification:  Suppose that $V$ is normed with norm $||\,||$. Can $V$ be given an inner product space structure such that $(x,x)=||x||^2$ ?
1217
Seems like I , I think I am missing something important.     Let $V$ be an $n$ dimensional inner product space ($n \geq 1$), and $T\colon\mathbf{V}\to\mathbf{V}$ be a linear transformation such that:         $T^2 = T$   $||T(a)|| \leq ||a||$ for every vector $a$ in $\mathbf{V}$;         Prove that a subspace $U \subseteq V$ exist, such that $T$ is the orthogonal projection on $U$.   Now, I know these things:   The fact hat $T^2 = T$ guarantees that $T$ is indeed a projection, so I need to prove that T is an orthogonal projection (I guess this is where $||T(a)|| \leq ||a||$ kicks in). To do this I can prove that:   For every $v$ in $ImT^{\perp}$, $T(v) = 0$ Alternatively, I can prove that for every $v$ in $ImT$ and $u$ in $KerT$, $(v,u)=0$. $T$ is  (according to Wikipedia) The matrix $A = [T]_{E}$ when $E$ is an orthonormal basis, is hermitian (this is equivalent to the previous point). What else?    I've been thinking about it for quite some time now, and I'm pretty sure there is something big I'm missing, again. I just don't know how to use the data to prove any of these things.  Thanks!
54217
Consider a vector space $V$ , show that: "$dimV=1\Leftrightarrow V$  has exactly two subspaces". Well I can understand that since $dimV=1$ the only subspaces are the zero and it self, but how can i prove that? How can I prove the equivalence? Thanks in advance.
111637
Suppose $(e_1,e_2,...,e_n)$ is an orthonormal basis of the inner product space $V$ and $v_1,v_2,...,v_n$ are vectors of $V$ such that $$||e_j-v_j||&lt; \frac{1}{\sqrt{n}}$$ for each $j \in \left\{1,2,...,n \right \}$. Prove that $(v_1,v_2,...,v_n)$ is a basis of $V$.  I am completely lost and just starting to learn about inner product spaces. Could someone provide a proof with the explanation of how you got there?
50184
Why is the span of the empty set defined to be $\{0\}$? It is known that the span of any nonempty set of vectors in a vector space $V$, gives a subspace of $V$, and it is stated in “Linear Algebra Done Right” by Axler, that to be consistent with this, the span of the empty set is defined to be $\{0\}$? Is this the only reason or does this definition prove useful in other ways later on?
51208
Let $A \in M_n(\mathbf{C})$ and let $B = \operatorname{adj}(A)$ be the adjugate matrix.      a) Prove that if $A$ is invertible, then so is $B$.   b) Prove that if $A$ has rank $n-1$, then $B$ has rank $1$.   c) Prove that if $A$ has rank at most $n-2$, then $B = O_n$.   a) We have that $A$ is invertible, so $\det(A) \neq 0$. The rank of $A$ is $n$. Also $$     A \operatorname{adj}(A)   = \operatorname{adj}(A) A   = \det(A) I. $$ But what next? Please help me to solve the problem using elementary theory  of matrix algebra (also please do not use the rank–nullity theorem).   EDIT: b) Clearly $\det(A) = 0$. Then $$     A \operatorname{adj}(A)   =  \operatorname{adj}(A) A   =  \det(A) I   =  0, $$ i.e. $A \operatorname{adj}(A) = 0$. But how does this imply that $\operatorname{adj}(A)$ has rank $1$?     May I request for any alternative solution of the problem? Can we conclude the result from $A \operatorname{adj}(A) = 0$?
102650
Let $T$ be a linear operator on a finite dimensional space $V$, and let $p=p_{1}^{r_{1}} \cdots p_{k}^{r_{k}} $ be the minimal polynomial for $T$, and let $V= W_{1} \oplus\cdots\oplus W_{k}$ be the primary decomposition for $T$, i.e., $W_{j}$ is the null space of $p_{j}(T)^{r_{j}}$. Let $W$ be any subspace of $V$ which is invariant under $T$. Prove that  $W= (W \cap W_{1})\oplus (W \cap W_{2})\oplus \cdots \oplus (W \cap W_{k})$.
57297
Let $A$ be an orthogonal $n × n$ matrix and suppose that $x, y \in \mathbb R^n$   and $\theta \in (0 , \pi)$ satisfy    $$Ax = \cos (θ)x − \sin (θ)y,\\ Ay = \sin (θ)x + \cos (θ) y.$$      Prove that $x$ and $y$ must be perpendicular vectors of equal length, i.e. show that $|x|^2 = |y|^2$   and $x · y = 0$.   We know that because $A$ is orthogonal then$$\langle x,x \rangle =\langle Ax,Ax \rangle, \quad \langle x,y \rangle=\langle Ax,Ay \rangle, \quad \langle y,y \rangle=\langle Ay,Ay \rangle.$$ I was already struggling with showing that $|x|^2 = |y|^2$. So far I just wrote out that   $$|x|^2= \sum_{i=1}^n (Ax_i)^2= \sum_{i=1}^n \cos^2(\theta)x_i^2-2\cos(\theta)\sin(\theta)x_iy_i+\sin^2(\theta)y_i^2,\\ |y|^2= \sum_{i=1}^n (Ay_i)^2= \sum_{i=1}^n \cos^2(\theta)y_i^2+2\cos(\theta)\sin(\theta)x_iy_i+\sin^2(\theta)x_i^2.$$  I don't really see how I could conclude from this that they are equal.
102739
Let $$A =\begin{bmatrix} 1&amp;3&amp;-6&amp;2\\5&amp;12&amp;-24&amp;7\\9&amp;15&amp;-30&amp;6\end{bmatrix}$$  $$RREF(A) =\begin{bmatrix} 1&amp;0&amp;0&amp;-1\\0&amp;1&amp;-2&amp;-1\\0&amp;0&amp;0&amp;0\end{bmatrix}$$ $$REF(A) =\begin{bmatrix} 1&amp;3&amp;-6&amp;2\\0&amp;1&amp;-2&amp;1\\0&amp;0&amp;0&amp;0\\\end{bmatrix}$$  Now, provided this information, it can be found that the basis for the nullspace of $A$ using the $RREF(A)$ is: $$\begin{pmatrix} \begin{bmatrix} 0\\2\\1\\0 \end{bmatrix}, \begin{bmatrix} 1\\-1\\0\\1 \end{bmatrix}\end{pmatrix}$$  Using the $REF(A)$ the "basis" for the nullspace is:  $$\begin{pmatrix} \begin{bmatrix} 6\\2\\1\\0 \end{bmatrix}, \begin{bmatrix} -2\\-1\\0\\1 \end{bmatrix}\end{pmatrix}$$  Is the "basis" that I provided for the $REF(A)$ actually a basis? How does the  $REF$ and $RREF$ of a matrix differ?
100804
If U is a T-invariant subspace of dim n+1, and if there is no subspace W ⊆≠U such that T(w) ∈ W for all w ∈ W, then the set of vectors {u,T(u),T^2(u),...,T^n(u)}, u ∈ U, u ≠ 0  forms a basis for U.   To show the set of vectors forms a basis. I need to show         The set is a linearly independent set    The span of this set is U.       I am stuck at this point. How can I show the vectors in this set is linearly independent? Also, I am not sure how to show the span(set) = U. I know the dimension of this set is also n, which matches to the dimension of the subspace. Can I claim since all the T-invariant subspaces are in U, the set of vectors span U? Thank you so much!    Edit: added u is non-zero
43037
This is a question from Finite-Dimensional Linear Algebra by Mark S. Gockenbach page 57 (exercise 2.5.15). I hope to check my proof.  Let $V$ be a vector space over a field $F$, and suppose $S$ and $T$ are subspaces of $V$ satisfying $S \cap T = \{0\}$. Suppose $\{s_1,s_2, ..., s_k\} \subset S$ and $\{t_1,t_2, ..., t_l\} \subset T$ are both linearly independent sets. Prove that $\{s_1,s_2, ..., s_k,t_1,t_2, ..., t_l \}$ is a linearly independent subset of $V$.  My proof) Suppose that $\alpha_1 s_1 +...+ \alpha_k s_k + \alpha_{k+1}t_1 +...+ \alpha_{k+l}t_l = 0$ for some $\{ \alpha_i\}_{i=1}^{k+l} \subset F$ Then $\alpha_{k+1}t_1 + ... + \alpha_{k+l} t_l = -(\alpha_1 s_1 + ... + \alpha_k s_k) \in S$ Since $\{ t_1,...t_l \} \subset T$ and $T$ is a subspace of $V$, $\alpha_{k+1}t_1 + ... + \alpha_{k+l} t_1 \in T $. Since $S \cap T = \{0 \}$, $\alpha_{k+1}t_1 + ... + \alpha_{k+l} t_l = 0$. Since $\{t_1, ... , t_l \}$ is a linearly independent set, $\alpha_{k+1} = ... = \alpha_{k+l} = 0$. By doing similarly, we can show that $\alpha_1 = ... = \alpha_k = 0$. Therefore, $\{s_1,s_2, ..., s_k,t_1,t_2, ..., t_l \}$ is a linearly independent subset of $V$. $\blacksquare$
108665
$\newcommand{\span}{\operatorname{span}}\newcommand{\rank}{\operatorname{rank}}$I would like to prove that for every martix, column rank = row rank Let $A\in M_{m\times n}$ be some matrix. fix $R_A$ = vector space of the rows of $A$ , $C_A$ = vector space of the columns of $A$  $$\rank (R_A) = dim \ (\span \{R_1, R_2, \ldots, R_m\})$$ $$\rank (C_A) = dim \ (\span \{C_1, C_2, \ldots, C_n\})$$  let $x$ be some vector. $$x\in \operatorname{Null}(A) \Leftrightarrow \forall i  \ \ (1 \leq i \leq m) : \langle x,R_i\rangle = 0$$ (to be clear - I'm referring to the inner product of $x$ with each row of $A$) Using the rank nullity theorem, $\dim \operatorname{Null}(A) =  n - \rank(R_A)$ as $n$ =  number of columns. what I would like to do is to say: $$\dim \operatorname{Null}(A) =  n - \rank(R_A)$$ $$\dim \operatorname{Null}(A) = n - \rank(C_A)$$ therefore,  $\dim \operatorname{Null}(A) = n - \rank(R_A)= n - \rank(C_A)  \Longrightarrow \rank(R_A) = \rank(C_A) $ it that false? is using the rank- nullity  theorem this way is cheating? or just doesn't prove what needs to be proven formally?
56555
I am trying to prove that given a subspace $\mathbf{W}$ in $\mathbb{R^{n}}$, the subspace and its orthogonal complement 'cover' whole of $\mathbb{R^{n}}$ through '+' where we define $\mathbf{W}$+$\mathbf{W^{\perp}}$ as linear combinations of vectors both in the subspace and in its orthogonal complement. It seems intuitively right, and I can prove that the sum of their dimensions adds up to n, but I am not sure how to prove the question I am looking at. Thanks!
108200
Let $V$ be a vector space. Given $U\le V$, define $U^\perp=\{f\in V^\ast\mid f(u)=0\,,\;\forall u\in U\}$ and given $W\le V^\ast$, define $W^\perp=\{u\in V\mid f(u)=0\;,\;\forall f\in W\}$. This defines a Galois connection between subspaces of $V$ and subspaces of $V^\ast$. If $V$ is finite dimensional, why is $\dim U+\dim U^\perp=\dim V$? Unfortunately, I am not very comfortable with linear algebra. Any hint will be appreciated.
96907
I am a bit unconfident with the following proof that I made to find the dimensions of the subspace below, $$ S =\{T\in\operatorname{Hom}(V, W): T(v)=0\} $$ where $V$ and $W$ are finite-dimensional vector spaces over F of dimensions $n$ and $m$ respectively and $v\in V$ be a non-zero vector. What I did is to define the following, $$ U:\operatorname{Hom}(V,W)\rightarrow W $$ where $T\mapsto T(v)$. And show $U$ is a linear transformation, then using the Dimension Theorem to find the $\operatorname{dim}(S)$ since $S=\operatorname{ker}(U)$. To show $U$ is linear is not so difficult, but I was confused at the end that was whether if $\operatorname{Im}(U)=W$. I know $\operatorname{Im}(U)\subset$ W, but if $\operatorname{Im}(U)\neq W$, then I can't derive the conclusion such that $\operatorname{dim}(S)= nm-m$. Additionally, I also have the following question that is to find the possible dimensions of the subspace below, $$ C =\{T\in\operatorname{Hom}(V,W):T(v_1)=0,T(v_2)=0,T(v_3)=0\} $$ where $v_1,v_2,v_3\in V$. Basically, I show this in a similar way to the previous one. Let $$ \Phi:\operatorname{Hom}(V, W)\rightarrow W\times W\times W $$ which is defined differently depending on the independence of $\{v_1, v_2, v_3\}$, but to show all of them are linear is not so difficult either. For example, if $v_1,v_2,v_3$ are linearly independent then $\Phi:T\mapsto (T(v_1),T(v_2),T(v_3))$. Since $C=\operatorname{ker}(\Phi)$ and if $\operatorname{Im}(\Phi)=W\times W\times W$, then using Dimension Theorem $\operatorname{dim}(C)=nm-3m$. Similarily, depending on the dependence between $v_1, v_2$ and $v_3$, I can also derive two possible dimensions that are $\operatorname{dim}(C)=nm-2m$ or $\operatorname{dim}(C)=nm-m$. I wonder how to end the proof (e.g. how to tell $\operatorname{Im}(U)=W$?), and whether if my guesses to the second question is correct or not. PS: I am a newcomer to linear algebra so if I made anything unclear or wrong in the proof, please help me to point it out since this is helpful for me to improve my skill when writing a proof. Thanks in advance.
14097
I am given that $V$ is n-dimensional vector space over $\mathbb{C}$ and $T \in L(V)$. And $T$ has least $m$ distinct nonzero eigenvalues. How do I show that $\text{null}(T^{n-m}) = \text{null}(T^{n-m+1})$?  I'm really not sure how to start here so just the first couple steps would be very helpful. Thanks!!