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5

Hint: The identity $TT^*=4T-3I$ implies that $T={1\over 4 }(TT^*+3I)$ You have $\langle T(x),x\rangle=\langle{1\over 4}(TT^*+3I)(x),x\rangle={1\over 4}(\langle TT^*(x)\rangle+\langle 3x,x\rangle)={1\over 4}(\langle T(x),T(x)\rangle+3\langle x,x\rangle)\geq 0 $. $(TT^*)^*=(4T^*-3I)=TT^*=4T-3I$ this implies that $T=T^*$. You have $T^2=4T-3I$ thus the ...


5

That equation is the definition of a self-adjoint linear operator, with respect to the scalar product. That is, $T$ is called self-adjoint if it satisfies that equation for all vectors $X,Y$. Certainly not all linear operators are self-adjoint, so you must have been missing a hypothesis that was made in the text you are citing.


3

For square matrices, $\underbrace{A^n=A\cdot A \cdot A \cdots A}_\text{n time}$ where $n \in \mathbb N$ So: $A^3 = A\cdot A \cdot A = $ $\begin{bmatrix}-2&1\\0&-1\end{bmatrix}$$\begin{bmatrix}-2&1\\0&-1\end{bmatrix}$$\begin{bmatrix}-2&1\\0&-1\end{bmatrix}$ The product operation of matrices is associative. you can evaluate $A^3$ ...


3

note that two similar matrix have the same caracteristic polinomial $P(t)$, if we calcul this polynomial for each matrix we tack an have coefficient -1 at $t^3$ and othere $0$ as coefficient at $t^3$ so the similarity is impossible


2

$4<x,y>= \vert\vert x + y\vert\vert ^2- \vert\vert x - y\vert\vert ^2$ Thus from your formula $4<x,y>= \sum _1^n \vert < x + y, e_i >\vert ^2- < x - y, e_i >\vert ^2$ and $<x,y>= \sum _1^n <x,e_i><y,e_i>= <x,\sum _1^n<y,e_i> e_i>$ By substraction $y-\sum _1^n<y,e_i> e_i$ is orthogonal to every $...


2

Let $A\in M_n(\mathbb{R})$; there are a symmetric $S$ and a skew-symmetric $K$ s.t. $A=S+K$; thus $tr(AX)=tr(SX)+tr(KX)=tr(SX)$ (since $X$ is symmetric, $tr(KX)=0$). We may assume that $S=diag(\lambda_1,\cdots,\lambda_p,\mu_1,\cdots,\mu_q,0_r)$ where $\lambda_i>0,\mu_j<0,p+q+r=n$. If $q>0$, then take $X=diag(0_p,xI_q,0_r)$ with $x>0$. Therefore ...


2

For the restriction $c+d=1$, rewrite this as $$cu+(1-c)v=v+c(u-v)$$ For $0\leq c$ this is a ray starting at $v$ and pointing towards $u$, as when $c=0$ the value is $v$ and we translate the vector in the direction of $u-v$ $1$ unit until we reach $u$, then continue to infinity. For $c<0$ we have a ray pointing in the opposite direction. For the ...


2

I think this is probably not possible in general. For a simple example, the matrix $$ \left(\begin{array}{cc}1 &1 \\ 0 & 1\end{array}\right) $$ has conjugates $$ \left(\begin{array}{cc}1 &x \\ 0 & 1\end{array}\right) $$ for $x \neq 0$. So you can make its Gershgorin radii arbitrarily small, but not zero...


1

The $T$ means transpose: $$u^T=\left[\begin{matrix}7 & 7 & 5\end{matrix}\right]$$ They're asking you to compute both $uu^T$ and $u^Tu$ because they give you very different results, which you'll see if you multiply these two matrices out. Hint: $uu^T$ is a 3-by-3 matrix and $u^Tu$ is a 1-by-1 matrix.


1

Working over $\mathbb C$, let's write $B$ as a block matrix $$ B = \begin{bmatrix}0 & X \\ E & 0 \end{bmatrix} $$ Then $$ B^2 = \begin{bmatrix}XE & 0 \\ 0 & EX\end{bmatrix} $$ and since $XE$ and $EX$ have the same eigenvalues, we can do a case analysis on the number of these eigenvalues: If there's only one eigenvalue for $EX$ and $XE$, ...


1

the condition $c+d=1$ is easy. take $$\vec{x}=c\vec{u}+d\vec{v}$$ now substitute $d\to1-c$, you obtain: $$ \vec{x}=c(\vec{u}-\vec{v})+\vec{v}\longrightarrow\lambda\,\vec{v_0}+\vec{v_1} \textrm{ (vectorial line equation)} $$ if you vary $c$ continously you obtain a line. for the other one you should look into Convex Combination (the explanation is longer and ...


1

MathLearner, to get such a map you need to make a choice of basis (directly or indirectly) at some point. In general, every subspace of a (finite dimensional) vector space naturally a kernel. Let $W\subset V$. Then $W$ is the kernel of the natural projection $\pi:V\to V/W$ (the quotient space). This is kind of 'the answer', since every map $\phi:V\to U$ ($...


1

Yes, this is true. Let $W$ be a subspace of $\mathbb{R}^n$, and let $B_W=\{w_1, \ldots, w_k \}$ be a basis of $W$. Complete $B_W$ to a basis of $V$, $B_V = \{w_1, \ldots, w_k ,v_{k+1},\ldots v_n\}$. Now define $T:\mathbb{R}^n \rightarrow \mathbb{R}^m$ as such: $\forall i, T(w_i) = 0$, and send $v_i$ to non-zero vectors. This uniquely defines a linear map ...


1

take a subspace $F$ of $\mathbb{R^n}$ then they exist $E$ subspace of $\mathbb{R^n}$ such that : $$ \mathbb{R^n}=E\oplus F $$ Take $P:\mathbb{R^n} \to \mathbb{R^n}$ the projection into $E$ then $P$ is linear and $$ Ker (P)=E $$


1

Row-reduction is equivalent to multiplying $M$ on the left by a series of invertible elementary matrices $A_1, \dots, A_k$. If $M$ is a nonsingular square matrix, its rref is $E_n$, so we have $A_k\cdots A_1M=E_n$. On the right side of the augmented matrix we have after row-reduction $A_k\cdots A_1E_n$ since the same operations were applied to both sides. ...


1

The answer is $\color{red}{no}$. If there is such a matrix and we consider the columns of $A$ as the vector $b$, you see that there are infinite inverses for matrix $A$ while we know the inverse of each matrix is unique. So, there is no such a matrix.


1

No, because if such matrix exist it will be surjective (by definition a surjective map is a maps who for all $B$ it exist a least one solution $X$ of $AX=B$) and because your matrix is square the matrix will by injective that mean $AX=0$ have only $0$ as solution. Contradiction


1

We have $(1)\ \Gamma(x_1,x_2,x_3)=(a x_1+(1-a) x_2, x_3)=(y_1,y_2)\rightarrow y_1=a x_1+(1-a) x_2\ \text{and} \ y_2=x_3.$ $(2)\ a x_1+(1-a) x_2+ x_3\leq 3 \ \rightarrow y_1+y_2\leq 3. $ $(3)\ x_i\geq 1,\ i=1,2,3 \ \rightarrow y_1=a x_1+(1-a) x_2\geq a+(1-a)=1 \ \ \text{and}\ \ y_2=x_3\geq 1 .$ So, according to $(1),\ (2),$ and $(3)$, it is concluded ...


1

The top left entry on the second matrix got copied wrong, it should be 1 not -1.


1

Definition: A set of vectors $\{\mathbf u_1, \dots, \mathbf u_k\}\subset \Bbb R^n$ is linearly independent if $$a_1\mathbf u_1 + \cdots + a_k\mathbf u_k = \mathbf 0 \implies a_1 = \cdots = a_k=0$$ So let's check if $\{(-6,2)\}$ is a linearly independent set. For we write down $\mathbf 0$ as a linear combination of the set: $$(0,0) = a(-6,2)$$ Are there ...


1

To difference them is important. For example, someone ask you find a max value in a space. Some other may put a constraint so it becomes to find a max value in a subspace. The two max values are different. Constraints are translated to subspace in math.


1

When you have free variables, then you can express the non-free variables in terms of them. So you will get some equation formed by the rows in RREF for instance: $x_1 - 2x_3 = 3$ $x_2 + x_3 = 1$ Now solve for x1 and x2... $x_1 = 3 + 2x_3$ $x_2 = 1 - x_3$ So now the vector $<x_1,x_2,x_3> = <3+2x_3,1-x_3,x_3> = <3,1,0> + <2,-1,1>...


1

Since that Q is orthogonal, $Q^Tv_1, Q^Tv_2,...,Q^Tv_p$ are orthonormal. Suppose that $\lambda_1,...,\lambda_k$ are zero eigenvalues of B and $k\leq p-1$, while $$\operatorname{tr}(V^TBV) = \sum_{i=1}^{n}\lambda_i\sum_{j=1}^{p}\tilde{v}_{ij}^2=0$$ Then we have $$\sum_{i=k+1}^{n}\lambda_i\sum_{j=1}^{p}\tilde{v}_{ij}^2=0$$ Thus $\tilde{v}_{ij}=0$ for $i=k+1,...


1

The most straight-forward way about solving this is to take the 2 equations and set them up as an Augmented Matrix and get RREF like so: $\left[\begin{array}{ccc|c} 1 & 1 & -3 & -2\\ 4 & 3 & 3 &2 \end{array}\right] \longrightarrow RREF \longrightarrow \left[\begin{array}{ccc|c} 1 & 0 & 12 & 8\\ 0 & -1 & 15 & ...


1

Suppose $$A=B+C$$ If $$B^T=B, $$ $$C^T=-C,$$ then according to the known property of transposition of sum of matrices $$A^T=(B+C)^T=B^T+C^T=B+(-C)=B-C$$ Now we have $$A=B+C \tag 1\\ $$ $$A^T=B-C\tag 2\\$$ Adding $(1)$ to $(2)$ gives $$B={(A+A^T)\over 2}\\ $$ Subtracting $(2)$ from $(1)$ gives $$C={(A-A^T)\over 2}\\ $$



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