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10

$A^2-4A-7I=0$ $\Longrightarrow$ $A^2-4A=7I$ $\Longrightarrow$ $A\cdot \frac17(A-4I)=I$. So $A$ is invertible. Its inverse is $\frac17(A-4I)$.

5

HINT: Let $v_1$, $v_2$ be column vectors of $I$ (identity matrix).

4

if $\lambda$ is an eigen value and the corresponding eigenvector is $A \neq 0,$ the $$A^\top = \lambda A$$ taking transpose gives you $$A = \lambda A^\top=\lambda^2 A$$ which show you $$\lambda^2 = 1\implies \lambda = \pm 1.$$

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 that $$A=P\begin{bmatrix}I_r& 0\\ 0 ... 2 It is not equal to the largest absolute value of eivenvalues. For a counterexample, consider a matrix A_{11}=0, A_{12}=A_{21}=A_{22}=1. However, it is a bound of eigenvalues. Let A be a complex square matrix. Then, all its eigenvalues are bounded by min \{ max_j \sum_i |A_{ij}|, max_i \sum_j |A_{ij}|\}. If you assume the matrix is symmetric, then ... 2 For a route different to induction, express the matrix as$$A=aI_3+G$$where I_3 is the identity matrix and G is given by$$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 ...

2

Let $\mathbf{v}_1=\begin{pmatrix}1\\0\end{pmatrix}$ and $\mathbf{v}_2=\begin{pmatrix}0\\1\end{pmatrix}$. By the condition given $\exists \, \mathbf{w_1}$ and $\mathbf{w_2}$ respectively, such that $A\mathbf{w_1}=\mathbf{v_1}$ and $A\mathbf{w_2}=\mathbf{v_2}$. So $$A\begin{bmatrix}\mathbf{w_1} & \mathbf{w_2}\end{bmatrix}=\begin{bmatrix}\mathbf{v_1} & ... 2 Observe that$$ \left(\frac17(A-4I)\right)A=I $$1 Hint: write I = AA^{-1}, so that:$$A^2-4A - 7AA^{-1} = 0.$$Can you solve for A^{-1}? 1 Hint: In terms of inner product, the projection of vector X=(x,y,z) onto vector u=(a,b,c) is simply:$$p_u(X)=\frac{\langle X,u\rangle}{\langle u,u\rangle}\, u$$For the second question, two strategies for the projection onto a plane: If you have the equation of the plane, as is the case, you also have a normal vector of the plane: n=(1,-2,3). ... 1 i think it is easier to find the projection on to the line u = (1, -2, 3)^\top that is orthogonal to the plane and then subtract from the identity to get the projection onto the plane. the projection matrix onto the line u is$$uu^\top/(u^\top u) = \frac1{14}\pmatrix{1&-2&3\\-2&4&-6\\3&-6&9}.$$therefore the projection matrix on ... 1 Call your vectors v_1,v_2. Pick two vectors v_3,v_4 such that \{ v_1,\dots,v_4 \} is a linearly independent set. Pick linearly independent images w_1,w_2 for them. (The linear independence ensures that the kernel contains only your given vectors.) Then you want A such that$$A \begin{bmatrix} v_1 & v_2 & v_3 & v_4 \end{bmatrix} = ...

1

Just follow your nose: Since $x=Ax+x-Ax$, take $u=Ax$ and $v=x-Ax$, and you only need to see that $v$ satisfies the required property: $Av=A(x-Ax)=Ax-A^{2}x=Ax-Ax=0$. Have you studied projections yet?

1

here is an almost algorithm to find a basis for the null space of a matrix $A:$ (a) row reduce $A,$ (b) identify the free and pivot b=variables. the variables corresponding to the non pivot columns, here they are $x_2$ and $x_4,$ are called the free variables and the rest are called pivot variables. (c) you can set one of the free variables to one and ...

1

\begin{cases} x_1+2x_2+\frac{23}{10}x_5=0 \\ x_3+\frac{13}{10}x_5=0 \\ x_4+\frac{22}{10}x_5=0 \end{cases} $$x_1=-2x_2-\dfrac{23}{10}x_5$$ $$x_3=-\dfrac{13}{10}x_5$$ $$x_4=-\dfrac{11}{5}x_5$$ Therefore,basis of null space= $$\begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\dfrac{23}{10} \\ 0 \\ -\dfrac{13}{10} ... 1 The null space of a matrix is basically a solution of the following: Ax = 0 x is a linear combination of all the independent matrices that satisfy the above equation. You have already multiplied the echelon form of A with x and have equated it to 0, so if any of the resulting equations is true, Ax = 0 We see that x5 is a variable in all the equations, so ... 1 Let's define the determinant a little differently than you may have seen before. Let's define it for linear functions f as:$$f(e_1) \wedge \cdots \wedge f(e_n) = \det(f)(e_1 \wedge \cdots \wedge e_n)$$where e_i is the i^{th} standard basis vector of \Bbb R^n. So first I'll have to go through what the exterior product, denoted \wedge, is. ... 1 An equation of the form a_1x_1 + \cdots a_nx_n = d defines an (n-1)-dimensional flat (affine subspace) in \Bbb R^n. Remember that it requires at least n of these (n-1)-D flats to uniquely define a point. Let's try to build some intuition for that. Consider \Bbb R^2. An (n-1)-D flat in \Bbb R^2 is a line. One line isn't enough to ... 1 They are linearly independent since the minor is non-zero, i.e :$$\text{det}\begin{pmatrix} 4 & -1 \\ -4 & 10 \end{pmatrix}=36\neq 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$$we can use this to write every power of A as a lineay combination of A, I. we will the division algorithm to find the remainder. suppose$$x^n = q(x)(x-a)(x-b) + \alpha x + ...

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) = ...

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) ...

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