Linked Questions

If a matrix $A$ has linearly independent columns, why is $A^TA$ invertible? I think it has to do with $A^TA$ being square and that $A^TAx$=0 only has the trivial solution, but I don't really know ...

I am trying to go in this direction only. Is this proof correct? ($A$ is an $m \times n$ matrix). I am considering the contrapositive of the statement: Suppose $A^TA$ is singular. Then $\det (A^TA)...

Say we have an $n\times m$ matrix $X$. What are the specific properties that $X$ must have so that $A=X^TX$ invertible? I know that when the rows and columns are independent, then matrix $A$ (which ...

Question: Prove that for a $m \times n$ matrix $A$, if $A^TA$ is invertible, then $A$ has linearly independent column vectors. I am hitting a complete blank with this proof, I have the following ...

I am learning Linear Algebra through Professor Gilbert Strangs lectures on MIT OCW. A concept I recently covered is finding the best solution to $ AX=b$ when $b$ does not lie in the column space of $...

I'm given a problem: $\sigma_1 \geq \sigma_2 \geq ... \geq \sigma_r$ are the nonzero singular values of $\textbf{A}\in\mathbb{R}^{M\times N}$. If $\epsilon \neq 0$ is a real scalar, s.t. $|\epsilon| &...

Consider the following matrix $ A = \begin{bmatrix} 0 & 4 & 4 \\ 1 & 1 & 1 \\ 4 & 0 & 4 \\ 4 & 4 & 0 \\ 1 & 1 & 1 \end{bmatrix}$ over $\mathbb{F}_5^{5 \times ...

Let $A\in \mathbb{R}^{m\times n}, m\ge n$ and $\operatorname{rank}(A) = n$. Prove that $A^tA$ is nonsingular I found this answer https://math.stackexchange.com/a/1840814/166180 which gives some ...

If $\mathcal B=${${x_1,x_2,...,x_n}$} and $\mathcal B'=${${y_1,y_2,...,y_n}$} are bases for an $n$-dimensional subspace of $\mathcal V \subset R^m$. Say I let $X_{mxn}$ and $Y_{mxn}$ be the matrices ...

Show that if $A \in \mathbb{C}^{m \times n}$ is of full rank, then null$(A^*)$ is the orthogonal complement of range$(A)$. I saw the above fact listed in this linked MSE proof. My attempt to ...

Suppose that the first column of $X \in R^{N \times (p+1)}$ is full of $1$s. Show that in the following cases, there exists no inverse matrices for $X^TX$. (a). $N < p+1$ (b). $N \ge p+1$ and two ...

If $X$ is a $N \times D$ matrix with $(D\gg N)$ with $\operatorname{rank}(X) = N$, what is $\operatorname{rank}(X^T \cdot X)$ where $X^T$ is the transpose matrix of $X$? I am little new to linear ...

Let $\mathbf{x_i}=[ x_{i1}, x_{i2}, x_{i3} ]'$. Why can we say that $\sum_i^n \mathbf{x_i}\mathbf{x_i}'$ is positive definite?