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6

Here is how I think about this: consider the case of a nilpotent operator $A$ on a vector space $V$ of dimension $n$, so that $A^n=0$. The characteristic polynomial is $T^n$. The $0$-eigenspace of $A$ is just its kernel; thus, in this case, the theorem says that the dimension of the kernel of $A$ is not more than the dimension of $V$, which is obvious. In ...

3

It is a classical result that you can always manage for the eigenvalues to be continuous, even when they are multiple. For hyperbolic matrices (i.e. with real eigenvalues) the result of Bronshtein ensures Lipschitz continuity for the eigenvalues. For your question in your framework the eigenvalue is smooth: with $p(X,t)$ be the characteristic polynomial, ...

3

I am trying to understand the question. Forgive me if I am totally off. But reflection in $y=x$ means geometrically that any vector on the line, stays where it is, which corresponds with an eigenvalue of 1. Any perpendicular vector to the line, is mapped exactly to its "other" side, which corresponds with an eigenvalue of -1. (The entries of the vector are ...

3

We are given: $$A = \begin{bmatrix}1 & 2 & 0\\2 & 1 & 0\\0 & 0 & 1\\\end{bmatrix}$$ The eigenvalues are found by solving $|A - \lambda I\ | = 0$, which yields: $$-\lambda^3 + 3 \lambda^2 + \lambda -3 = -(\lambda-3) (\lambda-1) (\lambda+1)$$ So, we have three distinct eigenvalues, which means we can diagonalize this system. To ...

3

Recall that $r$ is the smallest natural number $m$ such that $(A-\lambda I)^m=0_{\Bbb C^{n\times n}}$. The following holds: $$\begin{cases} (A-\lambda I)v_1=0_{\Bbb C^{n\times 1}}&\implies v_1\in \text{ker}(A-\lambda I) &\implies v_1\in \text{ker}\left((A-\lambda I)\right) \\ (A-\lambda I)v_2=v_1&\implies (A-\lambda I)^2v_2=(A-\lambda ... 3 Suppose (x_1,x_2)^{T} is an eigenvector, then$$\left[\begin{array}{cc}1&0\\4&1\end{array}\right]\left[\begin{array}{c}x_1\\x_2\end{array}\right]=\left[\begin{array}{c}x_1\\x_2\end{array}\right]$$The system of equation reduces to$$\begin{eqnarray*}x_1&=&x_1\\4x_1+x_2&=&x_2\end{eqnarray*}$$From the second equation we get x_1=0 ... 3 In general, any multiple of an eigenvector is also an eigenvector. Let A be any matrix and v an eigenvector with eigenvalue \lambda. Write w = \mu v. We show that w is an eigenvector for the eigenvalue \lambda as follows.$$ Aw = A(\mu v) = \mu Av = \mu \lambda v = \lambda w $$In fact, if the multiplicity of the eigenvalue \lambda is ... 2 Let \;\{v_1,...,v_n\}\; be a basis, or any lin. independent set, of \;V\; , then$$Tv_i=c_iv_i\;,\;\;c_i\in\Bbb F\;$$But also \;v_i+v_j\;,\;\;i\neq j\; , is an eigenvector so$$c_iv_i+c_jv_j=T(v_i+v_j)=c_{i+j}(v_i+v_j)$$Since the pair \;\{v_i,v_j\}\; is linearly independent, it must be that \;c_i=c_j=c_{i+j} . End now the argument. 2 We know that for every v\in S, Tv=c_vv, where c_v is some constant possibly depending on v. We want to show that c_v=c_w for all nonzero vectors v and w. It suffices to assume that v and w are linearly independent (if w=av, then clearly we must have c_w=c_v). For all such vectors$$c_vv+c_ww=T(v+w)=c_{v+w}v+c_{v+w}w.$$Rearranging, ... 2 Not knowing what exactly "canonical form" is, here is what I get. Translating to get rid of the linear terms:$$ 2(x+2)^2+4(x+2)(y-1/2)+6(y-1/2)^2=\frac{23}{2}\tag{1} $$With ... 2 Let, in general,$$x'=Ax+f\tag 1$$a system of n coupled ODE with x(t) and f(t) two n\times 1 vectors and A a n\times n matrix of constants. One way to solve these n coupled first order linear ODE is to diagonalize the coefficient matrix A and thus decouple these equations. Suppose that \lambda_1,\ldots,\lambda_n and v_1,\ldots, v_n are ... 2 If solving by decoupling, first find eigenvalues and corresponding eigenvectors. e1 = 1/2 ; e2 = 3$$v1 = \begin{bmatrix} 3 \\ 1 \end{bmatrix}v2 = \begin{bmatrix} 1 \\ -2 \end{bmatrix}P = \begin{bmatrix} 3 & 1 \\ 1 & -2 \end{bmatrix}P^-1 = \begin{bmatrix} 2/7 & 1/7 \\ 1/7 & -3/7 \end{bmatrix}P^-1*f(t) = \begin{bmatrix} ...

2

Your system can be reduced to $$\left\{ \begin{array}{ll} y+\beta z & =& 0, \\ (\beta\gamma-4)z &=& 0. \end{array} \right.$$ If $\beta\gamma \neq 4$ then $z=y=0$ and $x \in \mathbb{F}$. So, the eigenvectors are of the form $c\cdot(1,0,0)^T$, $c \in \mathbb{F}$. If $\beta\gamma = 4$ then system reduces to $y+\beta z = 0$ ...

2

Two vectors An eigenvector $v$ of a transformation $A$ is a vector that, when the transformation is applied to it, doesn't change its direction, i.e., $Av$ is colinear to $v$. The only thing that may change is its length. The factor of that change is its eigenvector $\lambda$. So, if $$Av_1 = \lambda_1 v_1, \quad Av_2 = \lambda_2 v_2, \quad \lambda_1 \ne ... 2 They are associated in R I'll concentrate on the eigen function from R, as you mention it in a comment. Quoting from its manual: If ‘r <- eigen(A)’, and ‘V <- rvectors; lam <- rvalues’, then A = V Lmbd V^(-1) (up to numerical fuzz), where Lmbd =‘diag(lam)’. This implies that the columns of ... 2 We are given:$$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\\ 1 & 0 & 0 & 1 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & 1 & 0 \end{pmatrix}$$We find that characteristic polynomial by solving |A - \lambda I| = 0, yielding:$$(\lambda -1)^2 (\lambda^2 + \lambda +1) = 0$$This yields a double and a complex conjugate pair of ... 2 Is in not the case that, whether A is rank-deficient or not, we have \langle x, Ay \rangle = \langle A^{\dagger} x, y \rangle = \langle Ax, y \rangle? \tag{1} And by (1), \langle x_k, \lambda_j x_j \rangle = \langle x_k, Ax_j \rangle = \langle Ax_k, x_j \rangle = \langle \lambda_k x_k, x_j \rangle, \tag{2} which immediately leads to \lambda_j ... 2 The answer is yes! In fact B is real analytic. (Not holomorphic, as you normalize.) Thus the Fourier series converges in any sense you want. Rank deficiency is irrelevant, as you can add a suitable multiple of the identity matrix, and pretty much nothing changes. The proof is not trivial. See Kato's Perturbation Theory. 2 Basically, the idea is in the previous answers. Nevertheless, I think the following proof is easier to understand. Let's prove the following: Fact: Suppose \lambda_0 is an eigenvalue of A and with geometric multiplicity k, then its algebraic multiplicity is at least k. Proof: By the assumption, we can find an orthonormal basis for the subspace ... 2 Elements of D are the eigenvalues, so they are the same as the diagonal elements of T. Columns of V are eigenvectors, but A = V D V^{-1}, called the Jordan decomposition, is numerically unstable, meaning that you can do it with the highest possible precision and still get A \ne V D V^{-1}. So, be sure that you really must use it. As for your ... 2 When T_1 or T_2 have repeated eigenvalues, the eigenvectors are not unique up to scalar multiplication. So, you cannot expect that your G is unique even if the eigenbases are required to be orthonormal. It is very easy to construct an example with multiple Gs. For instance, consider T_1=\operatorname{diag}(1,0) and T_2=I_2. Take u_1=(1,0)^T and ... 2 The geometry is what makes things easier (for me). Without the geometry, it would be a mechanical computation which I would not like doing, and might get wrong. Note that the vector (1,1) gets scaled by our two scalings to (2,2), and projection on y=x leaves it at (2,2). So the vector (1,1) is an eigenvector with eigenvalue 2. Now consider the ... 2 Eigenvectors solve the equation (A-\lambda I)x = 0. The quantity A-\lambda I is a matrix. Therefore, any solution to (A-\lambda I)x = 0 lies in the null space of A-\lambda I. The dimension of the null space of a square matrix is the size of the matrix minus is rank (this is the rank-nullity theorem, in a nutshell). If you look at the matrix above, ... 2 Hint: Since \dim(E_{\lambda _1})=n-1, there exist v_1, \ldots , v_{n-1} linearly independent eigenvectors of \lambda _1. Let v_n be an eigenvector of \lambda _2. Now consider the n\times n matrix P whose i^{\text{th}}column is v_i. Can you take it from here? 2 First, I think we all agree that this is obvious for T=0, so let's assume that T\neq0 for the rest of this post. First, let's look at the commutativity relation between T and U. Since TU=UT, if we pick an eigenvector v of U with eigenvalue \lambda, what can we say about UT\left(v\right)? When looking at this, just remember that ... 2 The proof is not conceptually difficult, it's just tedious to write down. Let A be an n\times n matrix. Let g be the geometric multiplicity corresponding to some eigenvalue \lambda of A and let m be the algebraic multiplicity. We will show that g \le m. Take your g linearly independent eigenvectors, say \{\mathbf{v}_1,\ \cdots,\ ... 2 Because a correlation matrix is symmetric , the right- and leftmultiplication to SVD/diagonalization are just the transposes of each other. Note, that "unique" does not mean "each one is different" here, but rather: we'll "unambiguously find a definitive solution" for the right- as well for the leftmultiplication. 2 1) Let me first answer what is "\vec{a} is in direction of vector \vec{b}". It means \exists c\neq0\in \mathbb{R}, that \vec{a}=c\vec{b}. It can also be written in this way$$ \cos(\theta)=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\cdot|\vec{b}|} =1 $$It is simpler in 2-d case, which is that m is the same for \vec{a} and \vec{b} if you express them ... 2 Note that U^T = U^{-1}, so the congruence$$A = U \Lambda U^T = U \Lambda U^{-1}$$is also a similarity relation. In other words, the eigenvalue decomposition is a unitary similarity of A and \Lambda. Since the same cannot be said about V := DU, relation (*) remains congruence that is not a similarity, hence it doesn't preserve the eigenvalues. ... 1 Looks to me like v_1=(3i,1)\implies \|v_1\|=\sqrt{10} so \displaystyle{v_1\over \|v_1\|}={1\over \sqrt{10}}(3i,1). And v_2=(-i,3)\implies \|v_2\|=\sqrt{10} so \displaystyle{v_2\over \|v_2\|}={1\over \sqrt{10}}(-i,3). Then$$U=\begin{bmatrix} v_1 & v_2\end{bmatrix}={1\over \sqrt{10}}\begin{bmatrix} 3i & -i\\ 1 & 3\end{bmatrix} and ...

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