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4

The set on invertible upper triangular matrices is actually closed in $GL(n,\Bbb R)$, since it is defined by the vanishing of a bunch of matrix entries (which entries are continuous functions of the matrix). If it were also open, it would be a union of connected components. But $GL(n,\Bbb R)$ has only two connected components (determined by the sign of the ...


3

Presumably, the author meant positive semidefinite, or specified something about the rank of $X$. Hint: Note that a matrix $A$ is positive semidefinite iff $v^TAv \geq 0$ for all vectors $v$. Note that $v^TX^TXv = (Xv)^T(Xv)$. As for symmetry: note that $(AB)^T = B^TA^T$.


2

Hint: Pick two rows (or two columns). Are they linearly independent?


2

$P=I - \frac{v v^T}{v^T v}$ is the orthogonal projection onto $v^\perp$. Proof: Clearly, $P$ is symmetric. $P^2 = (I - \frac{v v^T}{v^T v}) (I - \frac{v v^T}{v^T v}) = I - 2 \frac{v v^T}{v^T v} + \frac{v v^T}{v^T v} \frac{v v^T}{v^T v} = I - 2 \frac{v v^T}{v^T v} + \frac{v (v^T v) v^T}{(v^T v)^2} = I - 2 \frac{v v^T}{v^T v} + (v^T v) \frac{v v^T}{(v^T ...


2

One purely matrix-based definition of rank is decomposition rank: the rank of an $n\times m$-matrix is the smallest integer $r$ such that the matrix can be decomposed as product of an $n\times r$ and a $r\times m$ matrix. It is now obvious that the rank of $AB$ cannot be larger than the rank of $A$, or than the rank of $B$ (a decomposition of $A$ or of $B$ ...


2

The determinant is the same polynomial in the matrix entries no matter which field (or commutative ring) the entries come from. So what you're doing is right -- you can think of it either as doing the calculations in $\mathbb Z_5$, or as computing the determinant over $\mathbb Z$ and reducing modulo 5 at the very end.


2

In order to show that $(S + T) \in L(V,W)$, we need to show that $S+T$ satisfies the defining properties of $L(V,W)$. Clearly, $S + T$ takes a vector in $V$ and gives us a vector in $W$, so what we need to show is that $S + T$ is also linear. That is, we need to show that for $v_1,v_2 \in V$ and $k \in F$, we have: $$ (S + T)(v_1 + v_2) = (S + T)v_1 + (S ...


2

As $A$ and $B$ positive definite they have positive definite square roots: $$ A=A_1^2,\,\, B=B_1^2. $$ Clearly$^*$, $$ A-B\ge 0 \Longleftrightarrow B^{-1}_1AB_1^{-1} \ge I, $$ where $I$ is the unit matrix. Also$^{**}$, $$ B^{-1}_1AB_1^{-1} \ge I\Longleftrightarrow I \ge B_1A^{-1}B_1 \Longleftrightarrow B^{-1} \ge A^{-1}. $$ $^*$More specifically, if ...


2

This matrix say $A$ has the rank $n$ if $f$ is surjective i.e. $\operatorname{Im}f=\Bbb R^n$ and in the general case we see by the rank-nullity theorem that $$\operatorname{rank}A\le \min(n,m)$$


1

Let $x$, $y$, and $z$ denote the amount of swimmers who placed $1^{st}, 2^{nd}, \verb" and " 3^{rd}$ respectively. Translating the email into equations we get: 24 individuals placed $\Rightarrow x + y + z = 24$ (1) earning a combined total of 53 (3 points awarded for 1st, 2 for 2nd, 1 for 3rd). $\Rightarrow 3x + 2y +1z = 53$ (2) There were as many ...


1

One can easily generalize the answer $$ w = (1,-1,0,\ldots,0). $$ This vector is orthogonal to $v_1, \ldots v_n$ since all scalar products with are zero.


1

Knowing the definition of linear is central to the proof. Saying that the function $T : V \to W$ is linear means that $T$ has the following two properties: $T(cv) = cT(v)$ for all $c \in F$ and all $v \in V$ $T(v_1 + v_2) = T(v_1) + T(v_2)$ for all $v_1 \in V$ and all $v_2 \in V$ You first goal: Given two linear functions $S : V \to W$ and $T : V \to W$, ...


1

You're doing great so far. Your reduction is correct. To reduce the matrix even more, I would swap $R_{2}$ and $R_{3}$ so that you have a more obvious pivot in column 2 and then continue row operations. I would start by subtracting $R_{1}$ from $R_{2}$, so that you can zero out row(1), col(2) and then do $R_{3}$ + (3-b)$R_{1}$ to zero out row(3), col(2). ...


1

Hint: Let $x_1,\dots,x_n$ be a basis for $V$ over $\Bbb C$. Then $x_1,i\,x_1,x_2,i\,x_2,\dots,x_n,i\,x_n$ (where $i = \sqrt{-1}$) is a basis for $V$ over $\Bbb R$ (prove that this is the case).


1

Let me write the matrix in question in the form $L:=D-A$ (instead of $D+A$ to avoid alternating signs). Let $D$ be nonsingular and $\|D^{-1}A\|=:\epsilon<1$ for some operator matrix norm. We can write $$ L=D(I-D^{-1}A). $$ Using the Neumann series and with $B:=D^{-1}-D^{-1}AD^{-1}$, we have $$ ...


1

If $V$ is a nontrivial solution and $BA=I$, then $$ V = IV = (BA)V = B(AV) = BO = O, $$ which contradicts $V$ being nontrivial.


1

The simple Theorem to remember is that given a basis $\{\alpha_1, \alpha_2, ..., \alpha_n\}$ of $V$ and any $n$ vectors $\{\beta_1, \beta_2,.., \beta_n, \}$ in $W$ there is exactly one Linear Transformation such that $ T(\alpha_i) = \beta_i$. So you have two simple tasks: Find a basis $\{v_1, v_2\} $for $\ker T$ and let $T(v_1) = T(v_2) = \underline {0}$ ...


1

If $n$ vectors $v_1,..v_n$ are linearly independent(dependent) then for non singular matrix $P$ the vectors $Pv_1,...Pv_n$ also will be independent(dependent). From this fact and from the definition of the rank as a number of linearly independent columns (rows) we immediately can conclude that similar matrix have the same rank. The second fact follows ...


1

Matrix similarity: $\DeclareMathOperator{\rank}{rank}$ We say that two similar matrices $A,B$ are similar if $B = SAS^{-1}$ for some invertible matrix $S$. In order to show that $\rank(A) = \rank(B)$, it suffices to show that $\rank(AS) = \rank(SA) = \rank(A)$ for any invertible matrix $S$. To prove that $\rank (A) = \rank(SA)$: let $A$ have columns ...


1

No. You have described all the real symmetric matrices with nonzero determinant. The others are usually called semidefinite, for example $$ \left( \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right) $$


1

Ok, in English it is symmetric matrix. There must be some additional conditions you didn't tell. But if they are fulfilled: What did you do yourself? What is the transpose of $X^tX$? ( Wikipedia is your friend, if you don't know) What is for a vector $a$ then $Xa$ and what is $a^tX^t$ (if the not by you provided conditions are fulfilled)? From this you ...


1

If $X$ is $m\times n$, the best upper bound for $\Sigma_{\max}=\|X\|_2$ you can get is $\sqrt{n}$. The fact that this is an upper bound can be shown, e.g., by using the fact that $\|X\|_2^2\leq \rho(X^*X)\leq\mathrm{trace}(X^*X)=n$. The bound is attained for a matrix $X=[x,x,\ldots,x]$, where $\|x\|_2=1$. No reasonable upper bound for the minimal singular ...



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