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A more direct proof would be: Assume $\lambda$ is a non-zero eigenvalue of $f\circ g$, i.e. $$(f\circ g)v=\lambda v$$ for some $0\neq v\in V$. Application of $g$ on both sides shows $(g\circ f) (g(v)) =\lambda g(v)$ which shows that $\lambda$ is also an eigenvalue of $g\circ f$ (with eigenvetor $g(v)$, which is not equal to $0$, otherwise $(f\circ ... 1 This works not just for square matrices but all matrices where the product makes sense. A reference is Godsil and Royle's book Algebraic graph theory, Lemma 8.2.4. Their ingenious proof is to note that $$\det\left(\begin{bmatrix}I & A \\ B & I\end{bmatrix} \begin{bmatrix}I & 0 \\ -B & I\end{bmatrix}\right) = \det\left(\begin{bmatrix}I & ... 0 Depending what you mean by "find", this is straightforward if all eigenvalues are distinct, because then any matrix having those eigenvalues is diagonalizable http://en.wikipedia.org/wiki/Diagonalizable_matrix . In that case every invertible matrix P gives rise to a matrix Matrix_with_those_eigenvalues = P * diag([eigenvalues]) * inv(P) using MATLAB ... 1 Recall for linear PCA N is the number of data points in the set and comes in the eigen decomposition of the covariance matrix C, where C = \frac{1}{N} \sum_{i=1}^N x_i x_i^\top ( we look for eigenvectors v such that \lambda v = C v). So we simply multiply through by the constant N to bring it through to the other side. It is similar ... 3 Let A be m\times m matrix such that sum of each row is equal to n, then consider$$\begin{pmatrix}a_{11}&a_{12}&\cdots &a_{1m}\cr a_{21}&a_{22}&\cdots &a_{2m} \cr \vdots & \vdots & \ddots & \vdots \cr a_{m1}&a_{m2}&\cdots &a_{mm}\end{pmatrix}\begin{pmatrix} 1\cr 1 \cr \vdots \cr 1 ... 1 An Eigenvector is nothing more than a vector that points to some place No, an eigenvector is a vector that points to the same place after being transformed by the matrix. As far as usefulness of eigenvectors go, it is very hard to find a mathematical concept that has more real life applications then the concept of eigenvectors: for example, the ... 0 we will divide all the entries of the matrix by$\sqrt{4^2 + 5^2} = \sqrt{41}.$we can now write the matrix $$A = \pmatrix{ 4&-5\\5&4} = \sqrt{41}\pmatrix{ 4/\sqrt{41}&-5/\sqrt{41}\\5/\sqrt{41}&4/\sqrt{41}} = \sqrt{41}\pmatrix{\cos t&-\sin t\\\sin t &\cos t}$$ in a much simpler form. the matrix$A$represents a linear ... 0 Decomposition/ Determinant Method$A$can be decomposed into a matrix$S = sI_2$, where$I_2$is the$2\times 2$identity matrix, which scales the vector by a factor of$s$and a matrix$R_\theta$which rotates the vector by an angle$\theta$as$A=SR_\theta$. We know that$\det(A) = \det(SR_\theta) = \det(S)\det(R_\theta) = \det(S)$, because ... 0 Set$N = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}; \tag{1}$then a simple calculation reveals that$N^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix} \tag{2}$and$N^3 = 0. \tag{3}$Also,$A = aI + N. \tag{4}$Since$N$and$I$commute, we may apply the ... 2 For a route different to induction, express the matrix as $$A=aI_3+G$$ where$I_3$is the identity matrix and$Gis given by $$G=\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix}$$ A binomial expansion will yield \begin{align}A^k&=(aI_3+G)^k\\&={k \choose 0}a^kI_3+{k \choose 1}a^{k-1}I_3G+{k \choose ... 0 There is extended reading giving formula for calculating functions such as polynomials on matrixes: http://en.wikipedia.org/wiki/Matrix_function#Jordan_decomposition 3 I don't think the following could be caracterized as "nasty determinant calculation". I don't know how one can prove the equality without indulging in some computation. Let r=\operatorname{rank}(A) From a well-known theorem, derive that there exists P,Q invertible m\times m and n \times n matrices such thatA=P\begin{bmatrix}I_r& 0\\ 0 ... 1 i think you can do this with cayley-hamilton theorem which says that a matrix satisfies its characteristic equation. that is $$A^2 -(a+b)A + abI = 0$$ or $$A^2 = (a+b)A - abI$$ we can use this to write every power ofA$as a lineay combination of$A, I$. we will the division algorithm to find the remainder. suppose $$x^n = q(x)(x-a)(x-b) + \alpha x + ... 0 If this matrix is diagonalizable, that is if it has 4 distinct but not necessarily unique eigenvalues, then the eigenvalues are the entries in the diagonal matrix and the eigenvectors form the two adjacent matrices so we get PQP^{-1}. Even more basically the eigenvectors corresponding to eigenvalues are the vectors that when multiplied by the given matrix ... 1 Think of the matrix as a function from \Bbb R^4 to \Bbb R^4. If there is line through the origin in \Bbb R^4 that is taken to itself by the function, then it is an eigenvector and the amount that line is stretched or shrunk is the eigenvalue. It's easier to see this in \Bbb R^2. If the function is a rotation of 90^\circ then there are no (real) ... 1 We have$$A = X \begin{bmatrix}a & 0\\0 & b\end{bmatrix} X^{-1}$$This means$$A^n = X \begin{bmatrix}a^n & 0\\0 & b^n\end{bmatrix} X^{-1}$$We have$$A-bI = X \begin{bmatrix}a-b & 0\\0 & 0\end{bmatrix} X^{-1} \text{ and }A-aI = X \begin{bmatrix}0 & 0\\0 & b-a\end{bmatrix} X^{-1}$$Hence,$$\dfrac{a^n}{a-b}\left(A-bI\right) = ... 0 I don't see any need in invoking the Cayley-Hamilton theorem here. If$b_0\neq 0$, then it is invertible and will divide anything, even$0$(though in fact$\det A\neq0$in this case). If$b_0=0$then the minimal polynomial is divisible by$~X$, so$A$times a nonzero polynomial in$A$is zero, and$A$cannot be invertible; therefore$\det A=0$is a multiple ... 2 It is not equal to the largest absolute value of eivenvalues. For a counterexample, consider a matrix$A_{11}=0, A_{12}=A_{21}=A_{22}=1$. However, it is a bound of eigenvalues. Let$A$be a complex square matrix. Then, all its eigenvalues are bounded by$min \{ max_j \sum_i |A_{ij}|, max_i \sum_j |A_{ij}|\}$. If you assume the matrix is symmetric, then ... 5 Knowing just the eigenvalues of a matrix is not enough to tell whether it is magic. For example, the real matrices $$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \qquad \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$ both have the eigenvalues$1$and$-1$, but only the second of them is magic. So you need to adopt a more liberal ... 0 The eigendata refers to invariants of transformations. Different eigenvalues correspond to different eigenvectors/spaces, but not necessarily vice versa. 3 Hint:$D^{-1}=D$since$F=\mathbb{F}_3$and$D$is diagonal. 2 The system you propose is fine. You should find the following generalized eigenvectors: $$\vec{v}_3=\begin{bmatrix}0\\1\\0\\0\end{bmatrix},\quad\vec{v}_4=\begin{bmatrix}0\\0\\0\\1\end{bmatrix}$$ 2 First out, definition of Generalized Eigenvector, So you would want to try and solve$({\bf A}-\lambda {\bf I})^k{\bf v} = {\bf 0}$, but$({\bf A}-\lambda {\bf I})^{k-1}{\bf v} \neq {\bf 0}$to answer your question. You can use the first order eigenvectors (ordinary eigenvectors) together with the block-zero property to limit the span of generalized ... 0 Firstly, you can't show that$A$is diagonal just looking at ranks, but I think you mean diagonalizable. The fact that$A$has rank$5$means that it has an eigenvalue of$0$with geometric mulitiplicity$4$. Since$(A-3I)$has rank$5$, that means the eigenvalue$3$has geometric multiplicity$4$. You know that$5$is an eigenvalue, and it must have ... 0 i will show you how to find the eigenvector corresponding to the eigenvalue$1 + ih.$it follows the same lines as when the eigenvalue is real. you need to solve the homogeneous system of equations represented by the matrix $$\pmatrix{-ih&-1\\h^2&-ih}$$ we already know this matrix has rank <$2,$and in fact is of rank$1.$therefore the second ... 0 Use the exact same method as for real eigenvectors: $$\begin{pmatrix}1&-1\\h^2&1\end{pmatrix}\begin{pmatrix}x_i\\y_i\end{pmatrix}=\lambda_i\begin{pmatrix}x_i\\y_i\end{pmatrix}$$ So, for$\lambda_1=1+ih$: $$\begin{pmatrix}1&-1\\h^2&1\end{pmatrix}\begin{pmatrix}x_1\\y_1\end{pmatrix}=\begin{pmatrix}(1+ih)x_1\\(1+ih)y_1\end{pmatrix}$$ ... 2 Note that as$3$is odd, there must be at least one real eigenvalue. If there exists only$1$real eigenvalue, then that eigenvalue must be$\pm 1$because then the inverse of the eigenvalue also has to be an eigenvalue of$B$(by supposition). Otherwise there must exist$3$real eigenvalues, call them$x,y,z$. Then WLOG suppose$x=1/y$as then$x^{-1}$and ... 0 Hint: If$e$is an eigenvalue, its eigenspace is $$\ker(A-e\mathbb{I}_n)$$ Edit: Let's find the eigenspace of the second eigenvalue. This leads to finding the kernel of the matrix $$\left \lgroup\begin{matrix} ih & -1\cr h^2 & ih \end{matrix} \right \rgroup$$ If$h\neq 0$, we can multiply the second line by$\frac{i}{h}$, which yields $$\left ... 1 Check this out: Since A^3 - A+ I = 0, \tag{1} we have A(I - A^2) = (I - A^2)A = I. \tag{2} (2) shows that A^{-1} = I - A^2, \tag{3} so A is invertible. It is not in general true that (1) implies the characteristic polynomial of A is t^3 - t + 1; if \text{size}(A) \ne 3, for example, it cannot be the case, since the degree of the ... 17 Hint.$$ A^3-A+I=0\quad\Longrightarrow\quad A(I-A^2)=I. $$4 Call B= (-1)(A^2 - I), now (-1)(A^2 - I)A=I so you have B as one side inverse, on the other side also A(-1)(A^2-I) =I. 1 Suppose v_k is an eigenvector of A with eigenvalue \lambda_k so that Av_k=\lambda_kv_k then we have$$(A-\sigma I)v_k=\lambda_kv_k-\sigma v_k=(\lambda_k-\sigma)v_k$$Can you work out what the corresponding eigenvalue and eigenvector of A+\sigma I is? 1 Let \lambda be an eigenvalue of A with corresponding eigenvector X. We have$$AX=\lambda X.$$Consider$$B=A-a I.$$We have$$BX = (A-aI)X=AX-aIX=\lambda X-aX = (\lambda - a)X.$$Thus, if \lambda is an eigenvalue of A, then \lambda-a is an eigenvalue of B. 1 If c is an eigenvalue of A, then Av = c.v for the corresponding eigenvector. This implies (A - aI)v = (c - a)v. Thus, if c is an eigenvalue of A, then c-a is an eigenvalue of A - aI. This also shows that eigenvectors of A and A - aI are same. 1 Because A^2=-I, the minimal polynomial of A has to divide f(x)=x^2+1=(x+i)(x-i). So all of the possible eigenvalues for A are \{i\}, \{-i\}, or \{i,-i\}. 3 Hint If \lambda is an eigenvalue of A, then \lambda^2 is an eigenvalue of A^2. (Also, if A is real but you want to include complex eigenvalues, it's useful to know that if \lambda is an eigenvalue of a real matrix A, then (1) so is its complex conjugate \bar{\lambda}, and (2) \lambda and \bar{\lambda} have the same multiplicity.) 1 For any ring R, and any polynomials f(x), d(x), q(x) \in R[x], we have f(x) = d(x) q(x) \Rightarrow f_0 = d_0 q_0, \tag{1} where f_0 is the constant term of f(x) etc. This is easy to see by simply examining the basic rule for polynomial multiplication, with d(x) = \sum_0^{\deg d} d_i x^i, \tag{2} q(x) = \sum_0^{\deg q} q_i x^i, \tag{3} ... 2 The system you derived does not force the vector to be the zero vector: it only forces the second component to be zero. The first component can be anything to satisfy the system, and anything nonzero to be an eigenvector. For convenience you can choose the first component to be 1. Incidentally, because the system is triangular, you don't need to use ... 3 There exists a polynomial p(t) such that$$ \DeclareMathOperator{char}{char}\char_A(t)=p(t)m_A(t) $$It follows that$$ (-1)^n\det(A)=\char_A(0)=p(0)m_A(0)=p(0)b_0 $$That is,$$ \det(A)=(-1)^np(0)b_0 $$Hence b_0 divides \det(A). 4 Hint: For any square matrix of order n, the constant term of the characteristic polynomial \chi_Ais (-1)^n\det A, and the minimal polynomial of A divides \chi_A. 5 The space of polynomials is infinite-dimensional. So there is no characteristic polynomial, and no Cayley-Hamilton theorem. If g is a polynomial of degree n, then g(D) is the differentiation of order n. If you take a polynomial p of degree n+1, then g(D)p\ne0, hence g(D)\ne0. On the other hand, if you consider the space of polynomials of ... 1 The eigenvalues of A are the roots of its minimal polynomial, and \bigl\{P(x)\in \mathbf R[x] \mid P(A)=0\bigr\} is the set of multiples of the minimal polynomial of A. Hence if P(A)=0, the eigenvalues of A are among the roots of P(x). 2 Suppose v is an eigenvector of A, with eigenvalue \lambda. Then \lambda must satisfy the same cubic equation that A satisfies:$$(A^3-A^2-3A+2I)v=0v=0\\ (A^3-A^2-3A+2I)v=AAAv-AAv-3Av+2v\\ =AA\lambda v-A\lambda v-3\lambda v+2v\\ =\lambda A(Av)-\lambda(Av)-3\lambda v+2v\\ =\lambda A(\lambda v)-\lambda^2 v-3\lambda v+2v\\ ... 3 Hint : The minimal polynomial of the matrix$A$must be a divisor of the polynomial $$x^3-x^2-3x+2=(x-2)(x^2+x-1)$$ 0 This answer is an expanded version of my comment above. In short: your third computation is correct, your second one has a mistake, and your first one computes a different thing. First, let's make some things more precise. There are several subgroups of$\mathbb{S}^3\times \mathbb{S}^3$which are isomorphic to$C_4\times D_{12}^*$, so we must define the ... 0 Write$A=P\Lambda P^{-1}$where $$\Lambda=\DeclareMathOperator{diag}{diag}\diag(\beta_1i,\dotsc,\beta_ni)$$ with$\beta_j\in\Bbb R$. Then $$I+A=I+P\Lambda P^{-1}=P(I+\Lambda)P^{-1}=P\diag(1+\beta_1i,\dotsc,1+\beta_ni)P^{-1}$$ Thus the eigenvalues of$I+A$are$\lambda_k=1+\beta_ki$. In particular,$I+A$has nonzero eigenvalues. Hence$I+A$is ... 0 The eigenvalues of a skew symmetric (real) matrix are all purely imaginary: of the form$\;bi\;,\;\;0\neq b\in\Bbb R\;$, and they appear in conjugate pais, so we can list them as$\;bi\,,\,-bi\;,\;\;0\neq b\in\Bbb R\;$If$\;A\;$is diagonalizable and$\;J_A\;$is its JCF , then$\;J_A+I\;$has elements$\;1\pm bi\;,\;\;0\neq b\in\Bbb R\;$on its main ... 0 That's because$A$does not have eigenvalues -1. If it did,$\det(A+I)$would be 0. 2 Another side remark: You say that you are not sure if 1, or both 0 and 1 can be eigenvalues. In some cases, it is worthwhile to think of specific examples and see what they can tell us. So what are some examples of matrices$T$that satisfy$T^2 = T$? Well, the identity is certainly one, and its eigenvalues are all 1. However, another such matrix is the ... 7 Let$v\neq 0$be an eigenvector of$T$with eigenvalue$\lambda$, so$Tv=\lambda v$. Using$T=T^2$we have $$Tv = T^2 v = T(Tv) = T(\lambda v) = \lambda(Tv) = \lambda^2 v.$$ Hence,$\lambda v = \lambda^2 v$. Since$v\neq 0$we conclude$\lambda = \lambda^2$. The only solutions to this equation are$0$and$1\$.