# Tag Info

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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 ...

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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.

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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

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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 $$Ker (P)=E$$

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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 ...

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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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