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## Hot answers tagged matrices

8

If $M$ is diagonalizable, we have $M=PDP^{-1}$ with $D$ diagonal. Since $D^2=D$, we get $M^2=M$. Conversely, if $M^2=M$, $M$ is annihilated by $X^2-X=X(X-1)$ which has simple roots and splits over $\mathbb{F}_2$. So $M$ is diagonalizable. Hence the diagonalizable matrices are exactly the idempotents, i.e. $M$ such that $M^2=M$. This gives three ...

7

Starting with a $3\times 3$ matrix $$A=\left(\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right)$$ 1. Check that the matrix is invertible. First we need to check that we can actually invert the matrix. This amounts to showing that its determinant is non-zero. $$\det A=a\det\left(\begin{matrix}e & f \\ h & ... 5 Consider the possible dimension of the columnspace of the matrix BA. In particular, since A has at most a two-dimensional columnspace, BA has at most a two-dimensional columnspace. Stated more formally, if A has rank r_a and B has rank r_b, then BA has rank at most \min\{ r_a, r_b \}. 4 Joseph is right and  D^n=\begin{pmatrix}(-1)^n&0&0\\0&2^n&0\\0&0&3^n \end{pmatrix}so (Dt)^n=\begin{pmatrix}(-t)^n&0&0\\0&(2t)^n &0\\0&0&(3t)^n\end{pmatrix} so ... 4 Let$$A = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix}$$Then Ae_4 = -e_4. On the other side, the given three conditions and orthogonality of A imply Ae_4 \in \{\pm e_4\}, as Ae_4 \cdot Ae_i = e_4 \cdot e_i = 0 for i \in \{1,2,3\} hence Ae_4 \cdot e_j ... 4 Your professor is right, there's an infinite number of square roots, kind of like how there's two square roots of 1 (namely, 1 and -1). To see how to get it in general, notice that, for a symmetric matrix, you have$$ \begin{pmatrix}a&b\\b&c\end{pmatrix}^2 = \begin{pmatrix}a^2+b^2&b(a+c)\\b(a+c)&b^2+c^2\end{pmatrix} $$So, for the ... 4 Note that$$\mathrm{Tr}\left[(A-A^T)^2\right]=-2\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2.$$Therefore the identity can be rewritten as$$2(1-\mathrm{Tr}\,A)^2-\mathrm{Tr}\left[(A-A^T)^2\right]=8.$$Since \mathrm{Tr}\,AA^T=\mathrm{Tr}\,1=3, this is equivalent to$$\left(\mathrm{Tr}\,A\right)^2-\mathrm{Tr}\,A^2-2\mathrm{Tr}\,A=0.$$The left side of the last ... 4 Since the minimal polynomial has degree 4 which is the same order of the matrix, you know that A's smith normal form is \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & m_A(\lambda )\end{pmatrix}. Therefore the elementary divisors (I'm not sure this is the correct term in ... 3 Error 1: a-b=8 is not the same as a = \frac{8}{-b} but a = 8 + b Error 2: b+c=1 is not the same as b = \frac{1}{c} but b = 1-c Error 3: 3d+c=7 ... same as errors 1 and 2 Error 4: 2a-4d=6 Incorrect substitution because of previous errors. Fix the previous errors and then subsititue to fix this error 3 Hints: We are given the system:$$x' = Ax = \begin{bmatrix} 3 & -5\\ 5 & 3 \end{bmatrix}x$$with IC:$$x(0) = \begin{bmatrix} 2\\ -3 \end{bmatrix}$$The general solution will be:$$x_1(t) = c_2 \sin 4 t + c_1 (3 \sin 4 t + 4 \cos 4 t)x_2(t) = c_1 \sin 4 t + c_2 (4 \cos 4 t - 3 \sin 4 t)$$Using the IC, will yield the final solution ... 3 Use an augmented coefficient matrix, and obtain row-echelon form (using elementary row operations), to see if a solution exists, and/or if the system is inconsistent. If inconsistent, then no solution exists. 2x + 1y + 2z = A 0x + 2y + 2z = A 1x + 2y + 1z = A$$ M = \begin{pmatrix} 2 & 1 & 2 & A \\ 0 & 2 & 2 & A \\ 1 & ...

3

The characteristic polynomial is: $$|A - \lambda I| = 0 \rightarrow \lambda^2-6 \lambda+9 = 0 \rightarrow \lambda_{1,2} = 3$$ Substituting in the first eigenvalue to find the first eigenvector: $$[A - \lambda I]v_1 = 0 \rightarrow \begin{pmatrix} 2 & 4 \\-1 & -2\\\end{pmatrix}v_1 = 0$$ After RREF, for the first eigenvector, I would have ...

3

Steps to find Inverse 1. Find determinant of $3\times 3$ Matrix 2. Find minor 3. Find Cofactor 4. Find Adjoint 5. Replace results in below formula $A^{-1} = \frac{1}{\det(A)} adj(A)$ As you are unaware of these terms so let me first define it for you. If A is a square matrix, $(3\times 3)$ for example, then the minor of entry $a_{ij}$ , denoted by ...

3

if $rank(A^*A)=rank(A)$ it means $rank(A)=n-dim(null(A))$ and $rank(A^*A)=n-dim(null(A^*A))$ so $dim( null(A))=dim(null(A^*A))$ if exist $x$ such that $Ax\ne 0$ and $A^*Ax=0$ then $x\in null(A^*A)$but but x not belong to $null(A)$ and since $null(A)\subset rank(A^*A)$ so $$dim( null(A))\ne dim(null(A^*A))$$contradiction! on the other hand : we know ...

3

\begin{align*} 2x + 3y& = u\\ x - 4y &= v \end{align*}\quad \underset{\substack{\text{convert to}\\ \text{matrix language}}}{\leadsto}\quad \begin{bmatrix} 2 & \hphantom{-}3\\ 1 &-4\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix}u\\ v\end{bmatrix} \begin{align*} 3u - 5v &= c\\ 2u + 3v &= d \end{align*}\quad ... 3 Note that A has been diagonalized, and that:e^{At}=I+At+\frac{1}{2!}(At)^2+\frac{1}{3!}(At)^3+\cdots  \Longrightarrow e^{BDB^{-1}t}=I+BDB^{-1}t+\frac{1}{2!}(BDB^{-1}t)^2+\frac{1}{3!}(BDB^{-1}t)^3+\cdots  =BB^{-1}+B(Dt)B^{-1}+\frac{1}{2!}B(Dt)^2B^{-1}+\frac{1}{3!}B(Dt)^3B^{-1}+\cdots  ...

2

Joseph G. has it right. In addition, you say, "... I cannot compute eigenvalues" but the eigenvalues are simply the solutions to $\left|\begin{array}{ccc} -1-\lambda& 0& 0 \\ 0& 2-\lambda& 0 \\ 0& 0& 3-\lambda \end{array}\right|=0$ which boils down to $(-1-\lambda)(2-\lambda)(3-\lambda)$, so $\lambda=-1,2,3$.

2

Any real square matrix $A$ may be decomposed as $A = QR$ where $Q$ is an orthogonal matrix and $R$ is an upper triangular matrix (also called right triangular matrix). This generalizes to a complex square matrix $A$ and a unitary matrix $Q$. More generally, we can factor a complex $m\times n$ matrix $A$, with $m ≥ n$, as the product of an $m\times m$ ...

2

Note: My original answer contained an error; correcting the error led to this simpler solution. Most of the comments below refer to the original answer. You can project onto the subspace of vectors with constant entries, using a matrix with constant entries. Then $Ae_i$ is the same for all $e_i$, so the left-hand side of the desired inequality is $0$ unless ...

2

Hint: Consider first $$A^TA= \begin{bmatrix}l_1 & l_2 & l_3 \\m_1 & m_2 & m_3 \\n_1 & n_2 & n_3 \\ \end{bmatrix} \begin{bmatrix}l_1 & m_1 & n_1 \\l_2 & m_2 & n_2 \\l_3 & m_3 & n_3 \\ \end{bmatrix} = \cdots$$ by working out a few of the elements in terms of the components of $\mathbf {l,m,n}$. Then ask ...

2

This is not a complete answer to (a), but it may be a start. It also explains the repeated eigenvalues. It is known that a circulant matrix is diagonalized by the Fourier matrix and the eigenvalues of a circulant matrix are given by the discrete Fourier transform of its first row. See chapter 3 of here for a detailed explanation. Let $c_j$ be the first ...

2

Since (b) has been largely answered in the comments, I'll deal with (a) here. It is the case here that there is a very simple (and otherwise useful and well-known) basis of eigenvectors. Let $\zeta$ be a primitive $2n+1$-th root of unity. Our eigenvectors for $\Lambda$ will be the “cyclic” vectors, whose coordinates are successive powers of a root of unity : ...

2

Regarding your question about the inverse, of course $A^2=I$ tells you that $A=A^{-1}$. But for different $A$, no contradiction arises. Edit: in the first answer I didn't take care to produce symmetric matrices, and was using arbitrary invertible matrices. What one needs here is unitaries (i.e. orthogonal matrices). Here is one to construct all symmetric ...

2

For an easy constuction, consider matrices of the form $A=R_\theta^T\pmatrix{1\\ &-1}R_\theta$, where $R_\theta=\pmatrix{\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta}$ is a rotation matrix for angle $\theta\in[0,\pi)$. These $A$s are all distinct (exercise), real symmetric and their squares are equal to $I_2$.

2

Considering $\sigma(A)$ denotes the set of characteristic(eigen) values of A As $\lambda$ is an eigen value, $\exists 0\ne v\in V$ such that $Av=\lambda v$ Note that $A^kv=\lambda^k v,\forall k\in N$(easy to prove using Induction) Let $p(x)=\sum_{i=0}^{n}a_ix^i,a_i\in R$ Then we have , $p(A)=\sum_{i=0}^{n}a_iA^i$ So we have , ...

2

A sensible definition for this matrix would be a column-Latin rectangle, since the transpose is known as a row-Latin rectangle. Example: A. Drisko, Transversals in Row-Latin Rectangles, JCTA 81 (1998), 181-195. The $m=n$ case is referred to as a column-Latin square in the literature (this is in widespread use). I found one example of the use of ...

2

Hint: if you have $$Ax=\lambda x$$ then what do you get when you add $x$ to both sides? $$(I+A)x = (1+\lambda)x$$ Now do a bit of matrix algebra to get $(I+A)^{-1}$ on the right and $1+\lambda$ on the left. (Alternative way to think of it - if $(I+A)^{-1}x = \Lambda x$, then how can you rearrange this to get $Ax$ on its own?)

2

When you do the operation $4 R_2 - 5 R_1$, you are altering the determinant by a factor of $4$, since you are multiplying your matrix on the left by $$\begin{bmatrix}1 & 0 & 0\\-5&4&0\\0&0&1\end{bmatrix}.$$ Ditto for $3 R_1 + R_2$, which alters the determinant by a factor of $3$. These two operations have introduced the spurious ...

2

Consider similarity classes(equivalence class consisting of similar matrices) There are only three classes having representatives: $$\pmatrix{0& 0\\0&0}\;\pmatrix{1& 0\\0&1}\;\pmatrix{1& 0\\0&0},$$ respectively. Then consider conjugation with invertible matrices in $M_{2\times 2}(\mathbb{F}_2)$. There are only 6 of them. (i.e. ...

2

I'll answer this for $n\times n$ matrices, which gives only slightly more of a challenge. A diagonalisable matrix $A$ is completely determined by its eigenspaces, which form a direct sum giving the whole space, and of cuorse $A$ determines those subspaces as well. Over $\Bbb F_2$ only two eigenspaces are possible, for eigenvalues $0$ and $1$. The matrix $A$ ...

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