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4

The statement "$0 = a_01 + a_1z + \cdots + a_mz^m$ only holds when all $a_i = 0$" is a statement about polynomials being equal. You are talking about polynomials having a root. Certainly $z - z^2$ has zero as a root, so the equation $z - z^2 = 0$ has a solution, but the polynomial $z - z^2$ is not the zero polynomial, just like it's not equal to the ...


4

Suppose the diagonal entries of of $D$ are $\lambda_1,\lambda_2,\ldots, \lambda_n$. Let $p(x)$ be an interpolating polynomial such that $p(\lambda_k^3)=\lambda_k$ for each $k$. Then you can show that $p(A^3)=A$ (e.g. because this holds when restricted to an eigenbasis), and it is clear that $p(A^3)$ commutes with $B$. This relies on the fact that real ...


3

Yes. This is easy to see if you note that the diagonalization of $A$ produces $0$.


3

The linear transformation should satisfy that $$7\geq\underbrace{\dim(\text{Im}(f))=\dim \mathbf{R}^{10}-\dim(\text{Ker}(f))}_{\text{Rank-Nullity theorem}}=10-\dim(\text{Ker}(f))$$ Therefore $$\dim(\text{Ker}(f))\geq3.$$


3

Hint: $$\forall\,t\in [0,1]\;,\;\forall\,p,q\in B_1(\Bbb R^n)\;:\;\;\;||tp+(1-t)q||\le t||p||+(1-t)||q||$$ In fact, the above is true also for general $\;p,q\in\Bbb R^n\;$ , of course.


3

Yes true! To see this let $v=\sum_{i=1}^3 x_i e_i$ and $w=\sum_{j=1}^3 x'_je_j$ so $$b(v,w)=b\left(\sum_{i=1}^3 x_i e_i,\sum_{j=1}^3 x'_je_j\right)=\sum_{i=1}^3\sum_{j=1}^3x_ix'_j b(e_i,e_j)$$ so with $A=(b(e_i,e_j))_{1\le i,j\le3}$ we get $$b(v,w)=V^TAW$$ where $V=(x_1,x_2,x_3)^T$ and $W=(x'_1,x'_2,x'_3)^T$ the coordinates of $v$ and $w$ respectively.


2

Here is one proof: Note that $\|A\|_F^2 = \sum_k \|Ae_k\|_2^2$. Any set of orthonormal vectors will do, this also shows that $\|A\|_2 \le \|A\|_F$. Then $\|AB\|_F^2 = \sum_k \|ABe_k\|_2^2 \le \|A\|_2^2 \sum_k \|Be_k\|_2^2 = \|A\|_2^2 \|B\|_F^2 \le \|A\|_F^2 \|B\|_F^2$.


2

One way to do this would be to simply see what happens to your reference matrix if you start applying the matrices you have in your toolbox. By taking inverses appropriately (do you allow inverses in your product?), it is sufficient to write $\begin{pmatrix} s & 0 \\ 0 & s^{-1} \end{pmatrix}$ as a product of $\begin{pmatrix} a & b \\ c & d ...


2

Let us first "get rid" of the polynomials by introcducing coordinates. We have that $B=(1,X,X^2)$ is a basis. Since $f(1) = X^2 -1$, $f(X) = X^2 -k$, and $f(X^2) = X^2 -k^2$, we get that the matrix representing $f$ with respect to $B$ is $$M = \begin{pmatrix} -1 & -k & -k^2 \\ 0 & 0 & 0 \\ 1 & 1 & 1 \end{pmatrix}.$$ Computing the ...


2

Hint Look at the axioms. Obviously $0\in\mathcal S_2$ and $\lambda x\in\mathcal S_2$ for $\lambda\in\mathbb C$ and $x\in \mathcal S_2$. What will blow up is that $$x,y\in\mathcal S_2 \not\Rightarrow x+y\in\mathcal S_2$$ Try to find a counter-example for that, it's not hard ;)


2

That plane $V$ is described by $z=0$, hence your restriction is just $$q(x,y,0)=2x^2+3y^2+4xy$$ This is then $$q'(x,y)=2x^2+3y^2-4xy$$ which is represented by the matrix $$\begin{pmatrix} 2 & 2 \\ 2 & 3\end{pmatrix}$$ The eigenvalues are ${5\over 2}\pm {\sqrt{17}\over 2}$, which are both positive. so the new signature is $(2,0)$, in particular ...


2

The symmetric bilinear form that gives rise to your quadratic form $q$ is given by $$ b((x_1,y_1,z_1),(x_2,y_2,z_2))=x_1x_2-z_1z_2+2x_1y_2+2y_1x_2-y_1z_2-z_1y_2 $$ Now the conditions of being orthogonal to $(1,0,2)$ and to $(3,-1,0)$ are respectively given by setting for instance $(x_2,y_2,z_2)$ equal to that vector and equating the resulting expression ...


2

HINT: The eigenvalues of a real skew-symmetric matrix are pure imaginary and come in pairs of $(\lambda_k, - \lambda_k)$.


2

If the determinant $\delta$ exists, you can prove that $\delta(E)=c$ if $E$ is the elementary matrix corresponding to multiplication of a row by $c$; $\delta(E)=1$ if $E$ is the elementary matrix corresponding to summing a row to another multiplied by a constant; $\delta(E)=-1$ if $E$ is the elementary matrix corresponding to switching two rows; ...


2

$\Sigma$ has $K\cdot M$ rows and $L$ columns. Since $L\leq K,$ the statement "$\Sigma$ has full column rank" means $rank\ \Sigma = L.$ Put $$ \Xi := E(xz_1'), $$ i.e. $\Xi$ consists of the first $K$ rows of $\Sigma.$ By construction, $\Xi$ has $L$ rows. If we can show $rank\ \Xi = L,$ then the $L$ columns of $\Xi$ are linearly independent. This implies that ...


1

Such a problem is most effectively tackled by looking at each point and asking oneself How could this become wrong?. Let me get you started: Since $W$ is spanned by a set of three vectors, it has dimension at most $3$. Let's thus assume $W = \mathbb R^d$ with $d\in\{0,1,2,3\}$ for the tasks (this will actually be enough). For (A) to go wrong, $S$ must be ...


1

Denote $[A,B]:=AB-BA$. Hints: Calculate the following ones to recover the matrix $A$ of the desired equation $\delta(M)=[A,M]$: $$\left[\pmatrix{a&b\\c&d},\ \pmatrix{1&0\\0&0}\right],\quad\quad \left[\pmatrix{a&b\\c&d},\ \pmatrix{0&1\\0&0}\right],\quad \dots $$ Note that if $A$ is a solution (i.e. $\delta(M)=[A,M]$ for ...


1

The matrix $M_{WV}(\phi)$ of a linear transformation $\phi:E\to F$ in bases $V$ of $E$ and $W$ of $F$ is defined such that $$[\phi(x)]_W=M_{WV}(\phi)\cdot[x]_V$$ where $[x]_W$ denotes the coordinates of $x$ is the basis $V$, and $[\phi(x)]_W$ the coordinates of $\phi(x)$ in $W$. Then $$I\cdot[x]_W=[x]_W=[\phi\circ ...


1

There's an error in your formulation: a subspace $W$ of $V$ is $T$-invariant if $T(w)\in W$ for all $w\in W$. In other words: $W$ is $T$-invariant if $T(W) \subset W$.


1

I still don't really know what you are trying to say with definition 1, but I suspect you are trying to ensure some sort of uniqueness of coefficients, and apparently define the degree of a polynomial. I think you missed the mark here, so here is a simple and correct way to define a polynomial, and it's degree. Define a polynomial of degree $n\geq0$ to be ...


1

You can write the above expression in the following way $$\left(\begin{array}{lll}A_1\\A_2\\ \vdots \\A_N\end{array}\right)^T\left(\begin{array}{llll}B_1\\ & B_2\\ && \ddots\\ &&&B_N\end{array}\right)\left(\begin{array}{lll}C_1\\C_2\\ \vdots\\C_N\end{array}\right)$$ The Matrix in the middle is diagonal.


1

Yes, in fact, nothing has changed. What you have discovered is that the vector space $$\text{Mat}_{n\times m}(\Bbb R)\cong \Bbb R^{mn}$$ of matrices of size $m\times n$ with coefficients in $\Bbb R$ is isomorphic to an ordinary vector space, i.e. one of the form $\Bbb R^N$. You are just quoting the usual Minkowski inequality where, instead of labeling the ...


1

This is true if $P$ is invertible. Noting that $\operatorname{rank}(AB) \leq \min\{\operatorname{rank}(A),\operatorname{rank}(B)\}$, we have $\operatorname{rank}(AB) = \operatorname{rank}(BA) = \operatorname{rank}(A)$ whenever $B$ is invertible. Thus, $$ \operatorname{rank}(P^TAP) = \operatorname{rank}(P^TA) = \operatorname{rank}(A) $$


1

The space of skew-symmetric matrices of size $n$ and of rank $4$ has dimension less or equal than $n-3$, see here for a proof (using a classification result of skew-symmetric forms). For $n=4$ it follows that the space is $1$-dimensional. Hence the result.


1

show that it does not map the 0 function to the 0 function.


1

$T(0)$ is $T$ of the zero function which would give $T(0) = x \cdot 0 +1 = 0 + 1 = 1$. You don't even need to specify what function $f(x)$ is because it has been specified by writing $T(0)$.


1

Let $U$ be the diagonal matrix $U:={\rm diag}(u_1,u_2,\dots,u_n)$. Then, for a vector $a=(a_1,\dots,a_n)^T$ we have $Ua=(u_1a_1,\dots,u_na_n)^T$, and hence $$a^TUa= \sum_{i=1}^n u_ia_i^2\,.$$ Now apply it to $a:=Xw-y$. (If you want to arrive strictly at $w^Tx_i$, first note that $x_i^Tw=w^Tx_i$ as the inner product is commutative, then I think the $x_i$ ...


1

Let $\rho=\max_{\|x\|=1}\|Ax\|$ and $u=\arg\max_{\|x\|=1}\|Ax\|$. The case $\rho=0$ (i.e. $A=0$) is trivial. Suppose $\rho\ne0$. Then $Au=\rho v$ for some unit vector $v$ and $\|Av\|\le\rho$. Since $\langle u,Av\rangle=\rho$, we must have $Av=\rho u$. Therefore, $(-\rho,u)$ (when $v=-u$) or $(\rho,u+v)$ (when $u+v\ne0$) is an eigenpair of $A$. Normalize the ...


1

you want to find $(x, y, z)$ so that it is orthogonal to $(1,0, 2)$ and $(3, -1, 0)$ that means $x + 2z = 0, 3x - y = 0$ if the basis of is more complex you will make a matrix made of these vectors as rows and row reduce them to solve for $x,y,z.$ in this example it is easier and we can set $z = -1$ and solve for $x = 2, y = 6$ so that $V^\perp$ is spanned ...


1

You know that the matrix of the endomorphism is $$ A=\begin{bmatrix} 1 & 2 & a \\ 0 & 3 & b \\ -1 & 1 & c \end{bmatrix} $$ for some $a,b,c$. Elimination gives $$ \begin{bmatrix} 1 & 2 & a \\ 0 & 3 & b \\ 0 & 3 & c+a \end{bmatrix} $$ and, since we know that the rank must be $2$, we can conclude that $b=c+a$. We ...



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