Whenever we see a matrix $A=\bigl( \begin{smallmatrix} 3 &amp; 2 \\ 1 &amp; 2 \end{smallmatrix} \bigr)$ and $v=(3, 2),$ we can visualize that $(3, 2)$ represent the coordinates of $\mathbf i$ vector and $(1,2)$ represents the $\mathbf j$ vector. And to visualize $A$ as a whole, we can see it as a matrix that transforms can  the vector $v.$ when we multiply it. So, we can visualize $A^{-1}$ as a matrix that undoes the transformation by $A.$ Similarly, the determinant of $A$ gives the area. How can I visualize the transpose of a matrix as well as matrices that have a complex number as an element? Thank you.
53379
I can follow the definition of the transpose algebraically, i.e. as a reflection of a matrix across its diagonal, or in terms of dual spaces, but I lack any sort of geometric understanding of the transpose, or even symmetric matrices.  For example, if I have a linear transformation, say on the plane, my intuition is to visualize it as some linear distortion of the plane via scaling and rotation. I do not know how this distortion compares to the distortion that results from applying the transpose, or what one can say if the linear transformation is symmetric. Geometrically, why might we expect orthogonal matrices to be combinations of rotations and reflections?
22557
I need some visual intuition behind what exactly a symmetric matrix transformation does. In a $2 \times 2$ and $3 \times 3$ vector space, what are they generally?
122731
The cross product $\vec{a} \times \vec{b}$ can be written as the determinant of the matrix:  $$\left| \begin{matrix} \vec{i} &amp; \vec{j} &amp; \vec{k} \\ a_i &amp; a_j &amp; a_k \\ b_i &amp; b_j &amp; b_k \end{matrix}\right|$$  Is there any physical significance to this matrix, or is it just some mathematical trick?
21411
I have a tutorial problem to start with vector $$\begin{bmatrix}12\\5\\12\end{bmatrix}$$ and design a Householder transformation which annihilates the third component but leaves the first unchanged $$\begin{bmatrix}12\\?\\0\end{bmatrix}$$  I have looked at a few other answers but none quite seem to help me.  : This is pretty much the same question just different numbers. I get lost with the jargon in the accepted answer.  : This answer is on the same theme of what I am asking but the accepted answer is too vague for me.  I have these formulae: $\bf{v}=\bf{x} + sign(x_k)\lVert x\rVert \bf{e}_k$; $H=I-\frac{\bf{vv^T}}{\bf{v^Tv}}$  I have had many false starts on this that have led nowhere. Any ideas on where to start?
62077
Let $\Bbb R^2$ be a finite-dimensional vector space and $T:\mathbb{R}^2\to \mathbb{R}^2$ a linear operator. Let $[T]_B =  \begin {bmatrix}0 &amp; 0 \\-1 &amp; 1 \end{bmatrix} $ be a matrix representation of $T$ in the basis $B= (1,1),(-1,0)$.   How can I find the matrix $[T]_E$ where $E$ is the standard basis $(1,0),(0,1)$ if the transformation isn't given in the question?  Thanks,  Alan
39552
Use $\cal L$ to denote a linear transformation on some vector space. We know any matrix $\bf{A}$ can be viewed as a linear transformation by defining $\cal L:= \cal L(\bf{v})= Av$ where $\bf{v}$ is a vector. I am curious is any linear transformation can be represented by a matrix? If so, why? If not, can anyone help provide a counterexample?  In particular, projection of vectors onto a vector $\bf{b}$ is a linear transformation defined by ${\cal L(\bf{v})}=({\mathbf{v}} \cdot \frac{{\mathbf{b}}}{{|{\mathbf{b}}|}})\frac{{\mathbf{b}}}{{|{\mathbf{b}}|}}$. Can this linear transformation be represented by a matrix?  Thank you!
108668
Is it possible for a matrix $\mathbf{A}\in\mathbb{R}^{3\times3}$,  $$\mathbf{A}^2=-\mathbf{I}$$  I know that It is possible for $2\times2$ matrix, but is it possible for $3\times3$ matrix ?
48497
If $A$ is an $n \times n$ real matrix and       $$a_{ij}=\max(i,j)$$       for $i,j = 1,2,\dots,n$, calculate the determinant of $A$.   So, we know that   $$A=\begin{pmatrix} 1 &amp; 2 &amp; 3 &amp; \dots &amp; n\\  2 &amp; 2 &amp; 3 &amp; \dots &amp; n\\  3 &amp; 3 &amp; 3 &amp; \dots &amp; n\\  \vdots &amp; \vdots &amp; \vdots &amp; \ddots  &amp; \vdots\\   n&amp; n &amp; n &amp; \dots &amp; n \end{pmatrix}$$  but what do I do after?
7473
The following problem came up from a though I had while reading:  Let's say we have $M=\mathbb{Z}^n$ and we have another free $\mathbb{Z}$-module, $N$, inside of $M$ also with rank $n$.  We know we can make a matrix $A$ that changes $N$ to its representation in $M$ (ie has as columns the basis $N$ expressed in coordinates coming from the basis of $M$). If the index of $N$ in $M$ is $m$, I am pretty sure just from tinkering that the determinant of $A$ should also be $m$, however, I have not been able to show this.  It is unclear to me how to proceed. I though maybe I could do something related to the points in $\mathbb{Z}^n$ that are within the unit box of $N$, but I cannot really make sense of this.  Thank you for any help or direction.
58126
I'm about to use this diagram to make computing 3x3 determinants clear and easy for my students. I tried to add this picture to my lecture but it's blurred due to the bad quality of the picture. So i want to create exactly the same figure using latex.  I've searched and read a few topics about highlighting elements and diagonals in matrix but it does not work in this case. I want to add plus signs and minus signs as in the picture, and highlight elements in the matrice with elements outside the matrice.  I appreciate all your help and your advice. Thank you in advance !
64250
I am trying to understand how - exactly - I go about projecting a vector onto a subspace.  Now, I know enough about linear algebra to know about projections, dot products, spans, etc etc, so I am not sure if I am reading too much into this, or if this is something that I have missed.  For a class I am taking, the proff is saying that we take a vector, and 'simply project it onto a subspace', (where that subspace is formed from a set of orthogonal basis vectors).  Now, I know that a subspace is really, at the end of the day, just a set of vectors. (That satisfy properties ). I get that part - that its this set of vectors. So, how do I "project a vector on this subspace"?  Am I projecting my one vector, (lets call it a[n]) onto ALL the vectors in this subspace? (What if there is an infinite number of them?)  For further context, the proff was saying that lets say we found a set of basis vectors for a signal, (lets call them b[n] and c[n]) then we would project a[n] onto its . We project a[n] onto the signal-subspace formed by b[n] and c[n]. Well, how is this done exactly?..  Thanks in advance, let me know if I can clarify anything!  P.S. I appreciate your help, and I would really like for the clarification to this problem to be somewhat 'concrete' - for example, something that I can show for myself over MATLAB. Analogues using 2-D or 3-D space so that I can visualize what is going on would be very much appreciated as well.   Thanks again.
108962
I am trying to understand the Georgi chapter of tensor methods in $SU(3)$ representations, and I don't know how to resolve the tensor product of 2 matrices in a 2 heavy quark + 2 light antiquark system. Namely:  Let the tetraquark system be:           $QQ\bar q\bar{q}$  So, those $Q$ heavy quarks interact and can be understood in the $3\bigotimes 3$ representations for $SU(3)$, and $\bar q\bar q$ can be understood as the $\bar 3\bigotimes\bar 3=3\bigoplus\bar 6$ representations for $SU(3)$.  The interaction diagram for the $Q$ heavy quarks is shown in the following scheme, which is an analogy to the Feynman Diagram where a gluon is understood to be exchanged between upper and lower arrows:  $i'\xrightarrow{(T_{a})^{i'}_{i}}i ,  {}^iQ$  $j'\xrightarrow{(T_{a})^{j'}_{j}}j ,  {}^jQ$  So, as usuall we consider the elements of $3$ representation as $u^i$, and therefore the elements of $\bar 3$ representation as $v_{j}$.  I must understand how does the following expression transform:  $(T_{a})^{i'}_{i}(T_{a})^{j'}_{j}\cdot(1/2)\cdot (w^iv^j+w^jv^i)=\Xi\cdot(1/2)\cdot(w^{i'}v^{j'}+w^{j'}v^{i'})$  In other words, how can one give the correct value of $\Xi$?  Note1: $T_a$ is understood to be the $a-ith$ generator of the associated Lie Algebra, which come from the definition of the Gell-Mann Matrices.  $(T_{a})^{i}_{j}\propto(\lambda)_{ij}$, in matrix notation.  Note2: I understand the Young Tableaux and they are not intended to be used here, so please don't try to explain this situation through the tables.   Note3: I assume this is a purely solvable mathematical problem and doesn't need a deep insight into the physical concepts...  Also requesting information about how to understand the tensor products in these cases, since I have been checking many sources but the definitions seem not clear enough.
97848
Why is it, geometrically, that the row space is ALSO orthonormal? What exactly does the transpose LOOK like? Normally the row and column space are two separate things, but in the case of an orthonormal matrix, you have that the row space is actually the inverse of the column space, so also, what does an inverse space look like? I feel there should be a very good geometric interpretation, but I don't see it.
68464
I want to explicitly specify dimension of matrices in some expressions, something like   $$\boldsymbol{A}_{m \times n} \boldsymbol{B}_{n \times m} = \boldsymbol{C}_{m \times m} \, .$$  Is there any more or less standard notation for this?    While this notation is generally unambiguous, I think it become ambiguous, for example, in the following case. Suppose that I have some column or row vector which is conjugate transpose of correspondingly row or column vector. If I write it in the following way  $$\boldsymbol{A}_{n \times 1}^{\dagger} \, ,$$ we can have 2 different interpretations:   matrix $\boldsymbol{A}_{n \times 1}$ is $n \times 1$ column vector and by $\boldsymbol{A}_{n \times 1}^{\dagger}$ I'm referring to its conjugate transpose which is $1 \times n$ row vector; matrix $\boldsymbol{A}_{n \times 1}^{\dagger}$ itself is $n \times 1$ column vector which is conjugate transpose of $1 \times n$ row vector $\boldsymbol{A}$.
50986
This question regards the problem of . Vectors and matrices will be denoted in bold.  Premises. Let $N,K$ be given integers, with $K \gg N &gt; 1$. The transpose of matrix $\mathbf{A}$ will be denoted with $\mathbf{A}^T$. Suppose the following statistical model holds $$ (*) \quad \mathbf{y} = \mathbf{Hx + n}, \quad \mathbf{n} \sim \mathcal{N}_{K}(\mathbf{0}, \mathbf{C})  $$ where $\mathbf{y} \in \mathbb{R}^{K \times 1}$ are the observables, $\mathbf{H} \in \mathbb{R}^{K \times N}$ is a known full-rank matrix, $\mathbf{x} \in \mathbb{R}^{N \times 1}$ is a deterministic vector of unknown parameters (which we want to estimate) and finally $\mathbf{n} \in \mathbb{R}^{K \times 1}$ is a disturbance vector (noise) with a known (positive definite) covariance matrix $\mathbf{C} \in \mathbb{R}^{K \times K}$. The  (ML) estimate of $\mathbf{x}$, denoted with $\hat{\mathbf{x}}_{ML}$, is given by $$ (1) \quad \hat{\mathbf{x}}_{ML} = (\mathbf{H}^T \mathbf{C^{-1}} \mathbf{H})^{-1} \mathbf{H}^T \mathbf{C}^{-1} \mathbf{y} $$ and this is also the standard formula of Generalized Linear Least Squares (GLLS). Consider the standard formula of Ordinary Least Squares (OLS) for a linear model, i.e. $$ (2) \quad \hat{\mathbf{x}}_{OLS} = (\mathbf{H}^T \mathbf{H})^{-1} \mathbf{H}^T \mathbf{y} $$ As a final note on notation, $\mathbf{I}_K$ is the $K \times K$ identity matrix and $\mathbf{O}$ is a matrix of all zeros (with appropriate dimensions). Now, for the problem at hand, assume that $\mathbf{C}^{-1} = \mathbf{I}_K + \mathbf{X}$, where $\mathbf{X} \in \mathbb{R}^{K \times K}$ is a symmetric, invertible matrix (and, for the formalism to make sense, $\mathbf{X}$ is such that $\mathbf{I}_K + \mathbf{X}$ is invertible and the inverse is positive definite). In this case, it can be proven (using matrix inversion lemma) that $$ (3) \quad (\mathbf{H}^T \mathbf{C^{-1}} \mathbf{H})^{-1} \mathbf{H}^T \mathbf{C}^{-1} = (\mathbf{H}^T \mathbf{H})^{-1} \mathbf{H}^T + \mathbf{Q} $$ where the expression for the matrix $\mathbf{Q} \in \mathbb{R}^{N \times K}$ can be found (I will omit it here). Now, finally,  Proposition 1. If $\mathbf{H}^T\mathbf{X} = \mathbf{O}_{N,K}$, then equation $(1)$ degenerates in equation $(2)$, i.e., there exists no difference between GLLS and OLS.  The proof is straigthforward and is valid even if $\mathbf{X}$ is singular. Now, my question is  Question: Can an equation similar to eq. $(3)$ (which "separates" an OLS-term from a second term) be written when $\mathbf{X}$ is a singular matrix?  I am not interested in a closed-form of $\mathbf{Q}$ when $\mathbf{X}$ is singular. But I do am interested in understanding the concept beyond that expression: what is the actual role of $\mathbf{Q}$? In which space does it operate?  I found this problem during a numerical implementation where both OLS and GLLS performed roughly the same (the actual model is $(*)$), and I cannot understand why OLS is not strictly sub-optimal. I found this slightly counter-intuitive, since you know a lot more in GLLS (you know $\mathbf{C}$ and make full use of it, why OLS does not), but this is somehow "useless" if some conditions are met. What are these conditions?  As a final note, I am rather new to the world of Least Squares, since I generally work within a ML-framework (or MMSE in other cases) and never studied the deep aspects of GLLS vs OLS, since, in my case, they are just intermediate steps during the derivation of MLE for a given problem.  If the question is, in your opinion, a bit too broad, or if there is something I am missing, could you please point me in the right direction by giving me references? Preferably well-known books written in standard notation.
2545
(a) Given the column vector $a$ as an $n\times 1$ matrix. Write out its singular value decomposition, showing the matrices $U, \sum$, and $V$ explicitly.  (b) Given the row vector $a^T$ as an $1\times n$ matrix. Write out its singular value decomposition, showing the matrices $U, \sum$, and $V$ explicitly.  I currently get stucked on this question, since I couldn't find the eigenvalues of the $n\times n$ matrix $aa^T$ in part (a), and $a^Ta$ in part (b) to find the columns of $U$ and $V$, respectively. Can someone please help me on this problem?
49764
The Problem Let $A$ be the matrix $\bigl(\begin{smallmatrix}a&amp;b\\c&amp;d\end{smallmatrix} \bigr)$, where no one of $a,b,c,d$ is $0$. Let $B$ be a $2\times 2$ matrix such that $AB+BA=\bigl(\begin{smallmatrix} 0&amp;0\\ 0&amp;0 \end{smallmatrix} \bigr)$. Show that either   $a+d=0$, in which case the general solution for $B$ depends on 2 parameters, or $ad-bc=0$, in which case the general solution for $B$ depends on one parameter.   (this is question 22 of the last matrix exercise of Further Pure Mathematics by Bostock et al.)  Comments Writing $B=\bigl(\begin{smallmatrix} e&amp;f\\ g&amp;h \end{smallmatrix} \bigr)$ and multiplying out I get that    $(a+d)(f+g)+(b+c)(e+h)=0$ $ae+bg+cf+dh=0$   but I am unable to get the required restrictions on $a,b,c,d$. Is there a quick way of doing the problem that doesn't require manual computation? I thought of considering invertible and non-invertible cases but couldn't get anywhere. Help would be much appreciated.