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4

Hint. Write $$\frac{x_1}{3}=\frac{x_2}{4}=\frac{x_3}{2}=\lambda\ .$$ Then your set is $$W=\{\,(3\lambda,4\lambda,2\lambda)\mid \lambda\in{\Bbb F}\,\}\ .$$


3

Hint: By definition, we know that: $$ A(X_{\textsf{nullspace}}) = 0 $$ and: $$ A(X_{\textsf{particular}}) = b $$ But then by the linearity of matrix multiplication, we immediately get that...?


3

Here's a start: if you rewrite your equation as $A(A+I)+I = 0$, or $$ A[-(A+I)] = I $$ then you see that the quantity in brackets must be the inverse of $A$.


3

By Polarization identities (see http://en.wikipedia.org/wiki/Polarization_identity), we have: $$(Ax,Ay)=(x,y),\forall x,y\in \mathbb{C}^n,$$ which implies your result immediately.


3

Hint: Calculate $(SAS^*)^*$ using rules like $(AB)^*=B^*A^*$


2

It looks to me as if you've assumed that all the $v_i$'s (the $v_1,\dots,v_k$ in the problem and also the $v_{k+1},v_{k+2},\dots$ that came from nowhere) are orthogonal to each other. There's no justification for this assumption.


2

Here is another way of doing it: $$\eqalign{A^2+A+I=O\quad &\Rightarrow\quad (A-I)(A^2+A+I)=O\cr &\Rightarrow\quad A^3=I\cr &\Rightarrow\quad (\det A)^3=1\cr &\Rightarrow\quad \det A\ne0\cr &\Rightarrow\quad\hbox{$A$ is invertible}.\cr}$$


2

I guess that this could be shown by using the inequality $\|A\|_2\leq\sqrt{\|A\|_1\|A\|_{\infty}}$ and in a more general fashion. Let $A$ be such that $\|A\|_1\leq 1$, $b$ be a vector of unit 1-norm, and let $B:=\mathrm{diag}(b)$. We have $$ \|BA\|_2^2\leq\|BA\|_1\|BA\|_{\infty}. $$ Since the absolute row-sums of $A$ are bounded from above by one and the ...


2

The assertion applies not only to real tridiagonal matrices with positive sub and super diagonals, but also to every sign-symmetric real tridiagonal matrix whose sub or super diagonal entries are nonzero. What is missing in the paper you mentioned is the fact that every such tridiagonal matrix is diagonalisable (because $D^{-1}JD$ is real symmetric for some ...


1

The question amounts to finding when the matrices $$ A = \pmatrix{1\\&1\\&&1\\&&&b}, \quad B = \pmatrix{1\\&1\\&&0&c/2\\&&c/2&0} $$ are congruent. The answer in this case will be that the forms are equivalent iff $b < 0$ and $c \neq 0$.


1

A permutation $\sigma$ on a set $X = \{1,\dotsc,n\}$ is a bijection of the set $X$ on itself. Meaning that $\sigma$ maps any element of $X$ to another unique element of $X$ (maybe the same). For example, a permutation could look like this: $$ \sigma=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 5 & 4 & 3 & 1\end{pmatrix} $$ ...


1

Perron Frobenius theory says the largest eigenvalues is 1. I believe at this point, you just need the matrix to be normal, so that the spectral radius equals the operator norm.


1

Since the singular values of $A$ are the square roots of the positive eigenvalues of $A^tA$ then a necessary and sufficient is the eingevalues of $A^tA$ smaller or equal to 1. However, I believe this is not the answer that you want. So let me give you a sufficient condition which in some sense is sharp. A sufficient condition for the singular values of a ...


1

Those two vectors do not span the given subspace, and therefore cannot form a basis. Consider the vector $$\begin{pmatrix} 1\\ 1+1\\ 1\\ 1 \end{pmatrix}$$ which is in your subspace ($a=b=1$). This is not spanned by the two vectors, because any linear combination of them: $$c_1\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\ \end{pmatrix} + c_2\begin{pmatrix} 0 \\ 0 \\ 1 ...


1

To add to Peter's solution, If instead you were given $\mathcal{B}=\{(1,1,0,0),(0,1,1,1)\}$, then you have already shown that $(1,1,0,0)$ and $(0,1,1,1)$ span the subspace consisting of vectors of the form $(a,a+b,b,b)$. Call this vector subspace $V$. To show that $\mathcal{B}=\{(1,1,0,0),(0,1,1,1)\}$ is a basis it remains to show that $(1,1,0,0)$ and ...


1

Consider the point $(1,0,0)$. If $T(x,y,z)=(1,0,0)$ then $xy=0$ so $x=0$ or $y=0$. But then $y\sin x=0$, a contradiction. Hence $T$ is not onto. Note: this has nothing to do with linear algebra, as the function is nonlinear.


1

For the existence of a solution, the system of equations must be consistent. In this case, a solution exists when $$b_3 - 7b_1 - 2b_2 = 0.$$ Any value other than $0$ on the right-hand side would yield an inconsistent system (no solution exists, period), since the last row represents the equation $0x + 0y + 0z = b_3 - 7b_1 - 2b_2 = c$, and unless $c =0$, no ...


1

This is immediate, since both vectors belong to the $yz$-plane and are linearly independent (they're not scalar multiples of each other), and thus they form a basis of dimension $2$. If you want a more explicit proof, try solving for $c_1,c_2 \in \mathbb R$ in terms of $y,z \in \mathbb R$, where: $$ (0,y,z) = c_1(0,1,1) + c_2(0,2,-1) $$


1

I assume that all $v_i$ are non-zero. Here is one example for the assumption that $A$ is not linearly independent: $v_1 = v_2 + v_3$. In this case, obviously, $v_1 \in [ \{ v_2, v_3 \} ]$, i.e., $v_1 \not \in A^\perp$. Therefore, your first step is wrong. Further, I'm quite positive that the original statement is wrong. I think that $A^\perp = (A - ...


1

Try Matrix Analysis and Applied Linear Algebra by Carl Meyer. I think the book explains things very nicely and has many practical applications.


1

The bound on the degree is irrelevant. If $S$ is a set of polynomials such that $p(x)\ge 0$ for all $x$ and all $p\in S$, then $S$ is not a subspace so long as it contains a nonzero polynomial: for that polynomial you would have $p(x_0)>0$ for some $x_0$ and therefore $-p(x_0)<0$, so $-p\notin S$. So the only choice is $S=\{0\}$.


1

Hints: $\boxed{\Longrightarrow}$ Assume $X=Y^2$, for some symmetric matrix $Y$. Let $(\lambda ,v)$ be an eigenpair of $Y$. Since $Y$ is symmetric, what does that tell you about $\lambda$? Prove that $\left(\lambda ^2,v\right)$ is an eigenpair of $Y^2$. Conclude. $\boxed{\Longleftarrow}$ Assume all eigenvalues of $X$ are non-negative. Since $X$ is ...


1

If $Y^2 = X$ for some symmetric $Y$ then write $$ Y = P^TDP \\\implies X = P^TD^2P $$ the eigenvalues of $P^TD^2P$ are the values on the diagonal of $D^2$, that are squares of real numbers. Hence $X\ge 0$ (this is a notation for "$X$ has positive eigenvalues"). If $X$ has positive eigenvalues: You can write $E = \bigoplus_{i=1}^k E_{a_i}$ where $a_i\ge ...


1

A point $(1,1,1)$ is not a subset, but the set $\{(1,1,1)\}$ whose only element is that point, is a subset, and that might well be one of your sets. Similarly, $\mathbb R^3\setminus\{P\}$ where $P$ is any single point, would also be a subset that is not a subspace (why?)



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