# Tag Info

Your matrix $X$ is not orthogonal: the columns do not have norm $1$. Divide each column by its norm, and all will be fine then. Not being orthogonal means that $X^{-1}\neq X^T$.
To diagonalize, first notice that $$A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}3a\\3a\\3a\end{bmatrix}$$ so $$u:=\begin{bmatrix}1\\1\\1\end{bmatrix}$$ is an eigenvector of the $3a$ eigenvalue. To find eigenvectors $v,w$ of the $0$ eigenvalue, simply reduce $A$ to its RREF and determine a basis for the kernel. Then we can diagonalize $A$ as $A=P^{-1}... 2 Suppose$A$is a diagonalizable, and write$A=PDP^{-1}$. Then the equation becomes$PD^3P^{-1}=PD^2P^{-1}+PDP^{-1}-I$, which when we multiply by$P^{-1}$on the left and$P$on the right, we see is equivalent to$D^3=D^2+D-I$. Thus it suffices to show the diagonal case. Let$D$have diagonal entries$\lambda_i$($1\leq i\leq n$). Then for each$i$we have ... 2 Hint If$\lambda$is an eigenvalue, then $$\lambda^3-\lambda^2-\lambda+1=0 \Rightarrow (\lambda-1)^2(\lambda+1)=0 \Rightarrow \lambda=\pm 1$$ What is then$D^2$? Therefore $$A^2=PD^2P^{-1}=??$$ 2 Assume that we have$n$distinct eigenvalues for the$n\times n$matrix$A$then we can formulate the eigenvalue equations for all the eigenvalue and eigenvector pairs as $$Av_i=\lambda_iv_i \text{ for } i=1,2,...,n.$$ We can assemble all these equations into one single matrix equation $$A[v_1,...,v_n]=[v_1,...,v_n]\text{diag}[\lambda_1,...,\lambda_n]$$ ... 1 Guide: You just have to exhibit two matrices, one of which it is diagonalizable and one of which it is not. You might want to try matrix of this form. $$\begin{bmatrix} 2 & x \\ 0 & 2\end{bmatrix}$$ 1 If$M$is a symmetric matrix then we can choose eigenvectors$v_1,\ldots,v_N$of$M$such that$B:=\{v_1, \ldots, v_N\}$forms an orthonomal basis for$\mathbb{R}^N$. Let$C$be be the change of basis matrix which takes the standard orthonomal basis of$\mathbb{R}^N$to$B$. The matrix$C$has columns given by the eigenvectors$v_i$, that is $$C = \begin{... 1 Let g(x) be the minimal polynomial of f(A). Clearly, g has only simple roots, since f(A) is diagonalisable. Hence g and g' do not have any common roots, and since the eigenvalues of f(A) are roots of g, no root of g' is an eigenvalue of the matrix f(A), and therefore the matrix g'\big(f(A)\big) is invertible. Also, as f'(A) is also ... 1 Hints: 1) g’(f(A)) is invertible. 2) if B is a matrix, and p is a polynomial such that p(B)=0 and p’(B) is invertible, then B is diagonalizable. 1 If A is an m\times n matrix, then also \Sigma is m\times n, so U must be m\times m and V is n\times n. So if you want that U and V have the same shape, then A must be square. You're right in thinking that having U=V implies the (square) matrix A is normal: actually it must be Hermitian. If the SVD is A=U\Sigma U^\dagger, then A^... 1 For$$P^{−1}AP= Λ$$your matrix P is the matrix whose columns are the eigenvectors of A and Λ is the diagonal matrix of eigenvalues. Note that we need linearly independent eigenvectors in order for the matrix P to be invertible. 1 You have done everything correct except for the small mistake in the last step. While normalising the vector, you have multiplied by the norm instead of dividing it and hence you are getting different values. For the first column, instead of \frac{\sqrt{14}}{2} the term should be \frac{2}{\sqrt{14}}. And similarly for the other column. That should fix ... 1 You have correctly calculated two eigenvectors of H. In order to produce an orthonormal basis of eigenvectors (to get the corresponding unitary matrix), you want your eigenvectors v^{(1)}, v^{(2)} to satisfy:$$\left \langle v^{(1)}, v^{(2)} \right \rangle := v^{(1)}_1 \overline{v^{(2)}_1} + v^{(1)}_2 \overline{v^{(2)}_2} = 0 \ \ \ \text{ and}  \... 1 Notice that$A$satisfies the equation$x^3-x^2-x+1=0$. Upon factoring we see that the eigenvalues are$1$and$-1$, where$1$is of multiplicity$2$. So if$A$is diagonalizable, we can write$A$as$A=PDP^{-1}$, where the diagonal matrix$D\$ is given by \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & -1 \end{pmatrix} So, if we ...