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0

The matrix of $T$ is $A = \Big(\begin{array} &a & b \\ c &d \end{array} \Big)$. What you are given is $\Big(\begin{array} &a & b \\ c &d \end{array} \Big) \Big(\begin{array} &1 \\ 3 \end{array} \Big) = \Big(\begin{array} &-2 \\ -6 \end{array} \Big)$ and $\Big(\begin{array} &a & b \\ c &d \end{array} ...


2

Prove the following: $$\sigma\neq Id.\implies\;\;\exists\,i\;,\;\;1\le i\le n\;\;s.t.\;\;\sigma(i)>i$$ and from here that $\;a_{i\sigma(i)}=0\;$ ...


0

Standard matrix probably means the matrix of $T$ with respect to the standard basis $e_1 = (1,0)^T, e_2 = (0,1)^T$ of $\mathbb R^2$. I'd proceed by writing down the matrix of $T$ in the basis $(1,3)^T, (2,5)^T$ and then converting it into the standard basis. Let me know if this helps or if you want further instruction.


0

I think it probably assumes: Fathers are the same as mothers: So, - Mothers = fathers = 21 - Girls are 22 - Boys are 33 - Teachers are 16


0

The key idea that this exercise might be trying to teach you is that eigenvalues are independent of any basis you choose. So let's choose a basis, write down the matrix of $T$ in that basis and then compute the eigenvalues using the usual method. Why not choose the standard basis $b_1 = \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}, b_2 = ...


0

Let $T(x)$ be the linear transformation. Then there exists a unique A such that $T(x)=Ax$ for all x in $R^2$ In first example: $x = \Big(\begin{array} &1\\ 3\end{array} \Big)$. $T(x)=-2(x)$// Where x is a column vector. A standard matrix is the $m*n$ matrix whose jth column is the vector $T(ej)$, where ej is the jth column of the identity matrix in ...


0

This linear map is certainly invertible, so $0$ will not be an eigenvalue. Write down the 4 equations for $a,b,c,d$ you get when you say $T(X) = \lambda X$, for $X=\begin{bmatrix} a&b\\c&d \end{bmatrix}$. As a further hint, I will tell you that there are two eigenvalues. One eigenspace is $1$-dimensional and the other is $3$-dimensional.


0

An answer I found on Quora: It looks like ILA is a more introductory book, where LAaIA assumes that the reader is already familiar with the basics of matrices and vectors. ILA also seems to have some material introducing the abstract view of linear algebra, whereas LAaIA looks like it's mostly focusing on material that's relevant for engineering ...


0

If you look at row of people lined up in front of a mirror, not only will their mirror images appear in the opposite order (the image of the person closest to the mirror comes first) but also the image of each individual person will be a mirror image (if facing towards the mirror, the mirror image will be facing out of the mirror) just as if the person were ...


0

Given two matrices $M_{m\times n},N_{m\times p}$, there are two ways to interpret the entity $\begin{bmatrix} M & N \end{bmatrix}$. One is the $m\times (n+p)$ matrix whose $(i,j)$ entry is $\begin{cases} (M)_{(i,j)}, &\text{if }j\leq n\\ (N)_{(i, j-n)}, &\text{if }j\ge n+1\end{cases}$. In this case I'd rather denoted the matrix described above ...


0

The important concept here is linear dependence versus linear independence. As shown in the examples posted by others, linear dependence occurs when one equation in the system of equations can be shown to be a multiple of another. This is ultimately what Gaussian elimination or computing the determinant reveals. In this instance, there is no unique solution ...


0

a+b=c+d+e so teachers =13 monthers=24 father=18 boys=30 girls=25


0

$AB$ is similar to an SPD matrix, hence they have same eigenvalues: $$ AB\sim B^{1/2}(AB)B^{-1/2}\quad\Rightarrow\quad\lambda_{1}(AB)=\lambda_1(B^{1/2}AB^{1/2}). $$ Using this and the variational characterisation of $\lambda_1$, we have $$ \begin{split} \lambda_1(AB)&=\max_x\frac{x^TB^{1/2}AB^{1/2}x}{x^Tx} =\max_y\frac{y^TAy}{y^TB^{-1}y} ...


0

Let $C$ be a positive symmetric $n\times n$ matrix, and let $|\cdot|$ be the standard euclidean norm on $\Bbb R^n$. For all $X\in\Bbb R^n$, $$|CX|\leq\lambda_1(C)|X|$$ with equality iff $X$ is eigenvector of $C$ with eigenvalue $\lambda_1(C)$, so that, again, for $C$ be a positive symmetric, $$\lambda_1(C)=\max_{|X|\leq 1}|CX|=\|C\|$$ where $\|\cdot\|$ is ...


2

By back substitution, one can easily express $a,b,c,d$ in terms of $e$: \begin{cases} d=37-e,\\ c=e+7,\\ b=36-e,\\ a=e+19. \end{cases} Therefore $a+b+c+d+e=99+e$. It is very likely that they have simply forgotten the constraint that there are $100$ participants.


-1

Let the eigen values of $AB$ be $c_1, c_2, \ldots,c_n$ and those of $A$ and $B$ be $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ respectively. Now, we know that, $\det(A)$ is the product of the eigenvalues of $A$, counted with their algebraic multiplicities. Also, we have, $\det(AB) = \det(A)\det(B)$ $$ \Rightarrow c_1 c_2 \cdots c_n = (a_1 a_2 ...


0

Note that the element of the first line are the scalar products of the vector $(x_1, y_1)$ by the vectors $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, $(x_4, y_4)$.


1

Your matrix is nothing but $$\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\end{bmatrix} \begin{bmatrix} x_1& x_2& x_3& x_4\end{bmatrix} + \begin{bmatrix} y_1\\ y_2\\ y_3\\ y_4\end{bmatrix} \begin{bmatrix} y_1& y_2& y_3& y_4\end{bmatrix}$$ Hence, the rank of the matrix is $2$, in general. If $y_k = ax_k$ for all $k \in \{1,2,3,4\}$, where $a$ ...


1

Hints: $\;A\;$ is diagonalizable and we can choose a basis $\;B\;$ of $\;\Bbb R^2\;$ of its eigenvectors so that we'll have $$A(=[A]_B)=\begin{pmatrix}\frac23&0\\0&\frac95\end{pmatrix}$$ With respect to this same basis, we have for any $\;\binom xy\in\Bbb R^2\;$ : $$Ax=\binom{\frac23x}{\frac95y}$$ Now, just do some basic algebra with this. For ...


3

The following people take part in a school trip: 55 boys and girls; 43 girls and fathers; 42 fathers and mothers and 37 mothers and teachers. How many teachers took part in the school trips? Assuming the classes are mutually exclusive (i.e. no teachers are also parents) If you assume there were zero teachers, you get 37 mothers, 5 fathers, 38 girls and ...


4

We have: $$b=55-a$$ $$c=43-b=43-(55-a)=a-12$$ $$d=42-c=42-(a-12)=54-a$$ $$e=37-d=37-(54-a)=a-17$$ For all these numbers to be non-negative we therefore need $$17\le a\le 54$$with the inequalities strict if we require all the numbers to be positive. It was just possible that the constraints that all the numbers are non-negative (or positive) integers would ...


1

Your map $\lambda:V\times V\rightarrow V$ is continuous, but not bilinear: For $\mu\not=0\in \mathbb{R}$ and $v,w\not=0\in V$: $$\lambda(\mu v,w)=\mu v+ w\not = \mu (v+w)=\mu\cdot\lambda(v,w)$$ However, $\lambda$ is a linear map from the vector space $V\times V$ to $V$. Therefore it is continuous if and only if there exists $C>0$ such that ...


4

We add these equalities in this manner $$a+b+43+c+d+37=55+b+c+42+d+e$$ now we cancel we find $$\require{cancel}a+\cancel{b}+43+\cancel{c}+\cancel{d}+37=55+\cancel{b}+\cancel{c}+42+\cancel{d}+e$$ hence $$a+80=e+97\iff e=a-17$$ so each time you take a value of $a$ we find a value of $e$. Can you now answer your son?


-1

$R^n$ is if not said otherwise assumed to have a distance that derives from a norm. But all norms are equivalent in finite dimensional vector space (you can search for that in google), what you have in your question is half of the proof So using the infinite norm to prove this is like taking any other


0

The answer to your first question is negative (and hence there is no need to answer the second question). Consider $A=\pmatrix{1&1\\ 0&1\\ 0&0}$ and $C=\pmatrix{\frac{\sqrt{3}+1}2&\frac{\sqrt{3}-1}2\\ \frac{\sqrt{3}-1}2&\frac{\sqrt{3}+1}2\\ 0&0}$. Then $A^TA=C^TC=\pmatrix{2&1\\ 1&2}$. If $C$ was $UA$ for some unitary matrix ...


0

The answer is affirmative. Every positive semidefinite matrix $C$ can be orthogonally diagonalised as $QD^2Q^T$, where $Q$ is a real orthogonal matrix and $D$ is a nonnegative diagonal matrix. Let $\mathbf{Z}$ be a random vector following the standard multivariate normal distribution $N(0,I_n)$. It is straightforward to verify that $C$ is the covariance ...


0

By vectorising matrices in $M_n(\mathbb C)$ into vectors in $\mathbb C^{n^2}$, the inner product $\langle A,B\rangle=\operatorname{tr}(AB^\ast)$ is just the usual inner product $\langle \operatorname{vec}(A), \operatorname{vec}(B) \rangle = \operatorname{vec}(B)^\ast \operatorname{vec}(A)$ for vectors. So, the orthogonal complement in question is essentially ...


3

Denoting the matrix by $A$, you will find that multiplying yields $$A^2=A^3$$ (assuming my program is working, that is). This implies $A^2=A^3=A^4=\cdots$. If $A$ were diagonalizable, we could write $A=S\Lambda S^{-1}$ so that $$S\Lambda^2 S^{-1}=S\Lambda^3 S^{-1}=\cdots$$ $$\Lambda^2=\Lambda^3=\cdots$$ which means all the eigenvalues must be $0$ or $1$. ...


0

It can indeed. When you find the row space of a matrix, you are finding a basis for the subspace spanned by the row vectors. Likewise, when you find the column space of a matrix, you are finding a basis for the subspace spanned by the column vectors. Therefore, you have some freedom as to whether you want to plug the vectors into a matrix as its rows or ...


1

There are various methods of finding the matrix of $T$ with respect to a basis $B$. One of them is to work out one column at a time: if the basis is $\{{\bf b}_1,{\bf b}_2\}$ then you find the first column by calculating $T({\bf b}_1)$ and finding its coordinate vector with respect to $B$, and similarly for the other column(s). Here you have ...


2

If $A$ is invertible, then $B=A^{-1}C$, so $B$ is unique. Otherwise, $B$ can certainly fail to be unique. For example, let $A=\left(\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right)$ and let $C=\left(\begin{array}{cc}1 & 1 \\ 0 & 0\end{array}\right)$. Then $B$ can be any matrix of the form $B=\left(\begin{array}{cc}* & * \\ 1 & ...


1

Certainly there will be cases where, for some matrix $A$ and vector $b$, the equation $Ax = b$ has infinitely many solutions in vectors $x$. Now let $C$ be the matrix where each column is $b$.


0

Hint for subspace: let x and x' be in that set. You know that each of them have components that add up to zero, so if you add their components together, you must also get zero. You just need to rearrange the order to get it in the form of components of the element (x + x'). Scalar multiplication is a similar idea.


0

Your version of quadratic form is not enough. One normally includes the parallelogram law: $$ Q(\lambda x)=|\lambda|^{2}Q(x),\;\;\; Q(x+y)+Q(x-y)=2Q(x)+2Q(y), $$ where $x$, $y$ are vectors in a real or complex vector space $X$. There are subtle issues concerning quadratic forms on general linear spaces that are related to the axiom of choice; I ...


0

Let's denote $[v_k]_B = \sum_i \mu_{k,i}b_i$, and assume the $[v_k]_B$ do not form a basis. Obviously, at least one $\mu_{k,i}$ is non-zero for each $k$, and we would have a non-zero list $(\lambda_1, ..., \lambda_n)$ such that: $$ \sum_i\lambda_i \sum_j \mu_{i,j} b_j = \sum_j \left( \sum_i \lambda_i\mu_{i,j} \right) b_j = 0 $$ proving that $B$ is not a ...


0

Assuming that $\left[ \vec{x} \right]_B$ is the coordinate vector with respect to $\vec{x}$, here is a small hint. The map $\left[ \cdot \right]_B$ is linear, i.e., for $\alpha \in \mathbb{R}$ and $\vec{v} , \vec{w} \in \mathbb{R}^n$ $$ \left[ \alpha \vec{v} + \vec{w} \right]_B = \alpha \left[ \vec{v} \right]_B + \left[ \vec{w} \right]_B $$ This should get ...


2

Let's suppose $a,b,c$ are $\neq 0$. We have the system of equations in variables $\color{red}{\cos A}$, $\color{red}{\cos B}$ and $\color{red}{\cos C}$ \begin{array}{rcrcrcr} & & c\color{red}{\cos B} & + & b\color{red}{\cos C} & = & a \\ c\color{red}{\cos A} & & & + & ...


0

First, what is a symmetric matrix?, its definition already implies it has to be a nxn matrix. Second,nonsingular, same argument, it is nonsingular if and only if it has nonzero determinant, and only square matrices have nonzero determinant. Same for diagonal,only square matrices have diagonals,and same for d).


1

$V$ x $V$ consists of ordered pairs $(v_1,v_2)$. One can define a norm on $V $ x $V$ in various ways: For example a norm $||.||$ on $V$ x $V$ can be given by: $||(v_1,v_2)||=||v_1||+||v_2||$. Another norm could be: $||(v_1,v_2)||=max\{||v_1||,||v_2||\}$


1

Let the given points be $$ p_1 = \left[\begin{matrix} p_{11} \\ p_{12} \\ \vdots \\ p_{1n} \end{matrix}\right] \quad\text{,}\quad p_2 = \left[\begin{matrix} p_{21} \\ p_{22} \\ \vdots \\ p_{2n} \end{matrix}\right] \quad\text{,}\quad \dots \quad\text{,}\quad p_n = \left[\begin{matrix} p_{n1} \\ p_{n2} \\ \vdots \\ p_{nn} \end{matrix}\right] $$ You want to ...


1

Jim Hefferon has a freely available book Linear Algebra that discusses various applications as well as giving a solid theoretical base.


0

Strang revisited with a valid free download: https://archive.org/details/flooved1323 "Computational Sciences and Engineering - Applied Linear Algebra"


0

I'd try the following: $$\begin{cases}g_1=2f_1+f_2\\{}\\g_2=3f_3\end{cases}\implies g_1,g_2\in V$$ Since clearly $\;\dim V=2\;$ (why?) , it is then enough to prove $\;\{g_1,g_2\}\;$ is linearly independent, so suppose we have real scalars (I'm assuming the base field is the reals) $\;a,b\;$ s.t. $$0=ag_1+bg_2=2a\sin x+(a+3b)\cos x$$ The above is a ...


0

You're looking at a two dimensional space. Any two vectors that aren't on a line together (ie, aren't proportional to one another) will form a basis, and as you can plainly see by drawing a picture, $(2, 1)$ is not on a line with $(0, 3)$.


2

There is an innovative course Coding the Matrix offered by Philip Klein which consists of a book and a course offered on Coursera and other places. It even has a Twitter account for keeping updated. The reviews are controversial, see also here and here, but it looks as an interesting challenge to try. It is designed, according to the author's website, as a ...


3

Linear Algebra and its Applications- Gilbert Strang seems to be very recommended.


1

Linear Algebra and its Applications by David C. Lay is a simple book containing many references to real-world problems, including computer science.


1

Yes. If $EA=\mathbf{v}$ and $EB=\mathbf{w}$, then $AE=-\mathbf{v}$ and $BE=-\mathbf{w}$, so $$ AE \cdot BE = (-\mathbf{v}) \cdot (-\mathbf{w}) = (-1)^2 \, (\mathbf{v} \cdot \mathbf{w}) = \mathbf{v} \cdot \mathbf{w} = EA \cdot EB . $$


0

If you denote with $EA$ the vector with first end in $E$ and second end in $A$, then $$EA\cdot EB=AE\cdot BE\quad\forall A,B,E.$$ In fact, module of $EA$ is the same as $AE$, module of $EB$ is the same as $BE$ and the angle beetween $EA$ and $EB$ is the same as the one beetween $AE$ and $BE$ (the "sign" of the angle is also the same).


2

It is indeed true. It should be easy to confirm this by writing out the vectors explicitly in component form and evaluating both sides of the equality. Remember that if $x = \langle x_1, x_2, ... x_n \rangle$ and $y = \langle y_1, y_2, ... y_n \rangle$, then: $$x \cdot y = \sum_{i=1}^n x_iy_i$$



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