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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 ...

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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 $(AB)^T = B^TA^T$.

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

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$P=I - \frac{v v^T}{v^T v}$ 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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