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4

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 ...


3

No. It is quite possible for an eigenspace to have more than one dimension. As commenters above pointed out, examples include the identity matrix or the zero matrix.


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


3

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 $$ Ker (P)=E $$


2

We are given an initial state $x_0 \in \mathbb R^2$ and a final state $x_f \in \mathbb R^2$. We would like to find $A \in \mathbb R^{2 \times 2}$ such that $A^n x_0 = x_f$ for some $n \in \mathbb N$. Let $\Phi_n := A^n$. We then have a linear system in $\Phi_n$ $$\Phi_n x_0 = x_f$$ Vectorizing, we obtain an underdetermined linear system in a more standard ...


2

Trick: if an orthogonal matrix represent a rotation around some axis with amplitude $\theta$, such a matrix is similar to $$\begin{pmatrix}\cos \theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ but the trace of a matrix is left unchanged by matrix conjugation, hence in your case $$1+2\cos\theta = -\...


2

In practice you just check it by brute force. You can make it a bit faster to conduct by drawing a diagram with nodes indicating the elements 0, 1, 2, 3 and arrows between elements that are related (and such a diagram is good to do anyway in order to train your intuition). Then, instead of checking every combination of pairs for transitivity, you can just go ...


1

In case you have repeated eigenvalues, you won't have a unique eigenvector, but eigenspace that includes infinite number of vectors. BUT if your eigenvalue unique, then you have one eigenvector corresponding to it which up to a constant, by which u multiply this vector, i.e. [1 1] and [2 2] is the same.


1

As your question suggests, you can indeed use a matrix to visualize the relation. In your example, the matrix is $$ A = \left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 1\\ 1 & 0 & 1 & 1\\ 1 & 0 & 0 & 1 \end{array}\right]. $$ Note that there is a $1$ in the $(i,j)$-th cell if and only if $(i,j)\in R$. ...


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

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

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

If $A$ is $\geq 0$ and irreducible, then any eigenvector associated to the eigenvalue $\rho(A)$ has no zero entries. Here $x$ has a zero entry and consequently, $A$ is reducible. The answer is no. Indeed, consider the matrix $A=I_2$. It is reducible and $x=[1,1]^T>0$ is an eigenvector associated to $\rho(A)=1$.


1

If $X \in \mathbb R^{n \times n}$ is symmetric and positive definite, then there is a matrix $Y \in \mathbb R^{r \times n}$ such that $$X = Y^T Y$$ If $A \in \mathbb R^{n \times n}$ is also symmetric, then it has an eigendecomposition of the form $A = Q \Lambda Q^T$, where $Q$ is orthogonal. Hence, $$\mbox{tr} (A X) = \mbox{tr} (A Y^T Y) = \mbox{tr} (Y A ...



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