I have a 4x3 matrix and want to determine if the columns of my matrix are linearly independent. Is there a way to form a square matrix (4x4) in order get the ability to calculate the determinant and see if they are linearly independent?
15482
We can easily find out if the vectors are dependent or not by finding determinant of matrices formed from the vectors. But if the matrix is not a square matrix, we cannot find the determinant. Then how do we know if the vectors are linearly independent?
33605
\begin{array}{rrrrr|r}     b &amp; a &amp; a &amp; \cdot \cdot \cdot &amp; a \\     a &amp; b &amp; a &amp; \cdot \cdot \cdot &amp; a \\     a &amp; a &amp; b &amp; \cdot \cdot \cdot &amp; a \\     \cdot &amp; \cdot &amp; \cdot &amp; \space &amp; \cdot\\ \cdot &amp; \cdot &amp; \cdot &amp; \space &amp; \cdot\\ a &amp; a &amp; a &amp; \cdot \cdot  \cdot &amp; b   \end{array}  I have the above matrix $A\in M_{n\times n}(F)$ where $F$ is a field and $n\geq1$, $a,b\in F$.  I'm trying to find out how to use row operations to make it into an upper triangular matrix in order to figure out the determinant. But I'm not sure how I would approach it.
101079
Let $A$ be an $n\times n$ matrix.  If $A$ is invertible then one of it's eigenvalues is 0.  If $A$ is diagonalizable then it has $n$ linearly independent eigenvectors.   Are these two statements true? Is there any connection between them? May I infer one from the other?
101698
Could anyone just give me hint for this one?  There exist an infinite subset $S\subseteq\mathbb{R}^3$ such that any three vectors in $S$ are linearly independent. True or false?
57100
When can we split a square matrix (rows = columns) into it’s LU decomposition?  The LUP (LU Decomposition with pivoting) always exists; however, a true LU decomposition does not always exist.  How do we tell if it does/doesn't exist?  (Note: decomposition and factorization are equivalent in this article)  From the :     Any square matrix $A$ admits an LUP factorization. If $A$ is invertible, then it admits an LU (or LDU) factorization if and only if all its leading principal minors are non-zero. If $A$ is a singular matrix of rank $k$, then it admits an LU factorization if the first $k$ leading principal minors are non-zero, although the converse is not true.   This implies that for a square matrix:     LUP always exists (We can use this to quickly figure out the determinant). If the matrix is invertible (the determinant is not 0), then a pure LU decomposition exists only if the leading principal minors are not 0. If the matrix is not invertible (the determinant is 0), then we can't know if there is a pure LU decomposition.     The problem is this third statement here. “If $A$ is a singular matrix of rank $k$, then it admits an LU factorization if the first $k$ leading principal minors are non-zero”, gives us a way to find out if LU decomposition exists for a singular (non-invertible) matrix.  However, it then says, “although the converse is not true”, implying that even if a leading principal minor is 0, that we could still have a valid LU decomposition that we can't detect.  This leads us back to the question: is there a way of truly knowing whether a matrix has an LU decomposition?
44409
I know that given a $2N \times 2N$ block matrix with $N \times N$ blocks like   $$\mathbf{S} = \begin{pmatrix} A &amp; B\\  C &amp; D \end{pmatrix}$$  we can calculate $$\det(\mathbf{S})=\det(AD-BD^{-1}CD)$$ and so clearly if $D$ and $C$ commute this reduces to $\det(AD-BC)$, which is a very nice property.   My question is, for a general $nN\times nN$ matrix with $N\times N$ blocks where all of the blocks commute with each other, can we find the determinant in a similar way? That is, by first finding the determinant treating the blocks like scalars and then taking the determinant of the resulting $N\times N$ matrix.  Thanks in advance!
30552
I know that I can find a basis for the column space of a matrix $A$ by reducing the matrix to reduced row echelon form $J$. The columns of $A$ corresponding to the linearly independent columns of $J$ then form a basis for $Col(A)$, because linear dependence is preserved under elementary row operations. I can't figure out why this is true though, and a google search returns nothing, so I'm sure it's simple. Can someone give me a proof?
34167
In my linear algebra class, we just talked about determinants. So far I’ve been understanding the material okay, but now I’m very confused. I get that when the determinant is zero, the matrix doesn’t have an inverse. I can find the determinant of a $2\times 2$ matrix by the formula. Our teacher showed us how to compute the determinant of an $n \times n$ matrix by breaking it up into the determinants of smaller matrices. Apparently there is a way by summing over a bunch of permutations. But the notation is really hard for me and I don’t really know what’s going on with them anymore. Can someone help me figure out what a determinant is, intuitively, and how all those definitions of it are related?
10950
I have the following $n\times n$ matrix:  $$A=\begin{bmatrix} a &amp; b &amp; \ldots &amp; b\\ b &amp; a &amp; \ldots &amp; b\\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots\\ b &amp; b &amp; \ldots &amp; a\end{bmatrix}$$  where $0 &lt; b &lt; a$.  I am interested in the expression for the determinant $\det[A]$ in terms of $a$, $b$ and $n$.  This seems like a trivial problem, as the matrix $A$ has such a nice structure, but my linear algebra skills are pretty rusty and I can't figure it out.  Any help would be appreciated.
13830
A Vandermonde-matrix is a matrix of this form:  $$\begin{pmatrix}  x_0^0 &amp; \cdots &amp; x_0^n \\ \vdots &amp; \ddots &amp; \vdots \\  x_n^0 &amp; \cdots &amp; x_n^n \end{pmatrix} \in \mathbb{R}^{(n+1) \times (n+1)}$$.  condition ☀ : $\forall i, j\in \{0, \dots, n\}: i\neq j \Rightarrow x_i \neq x_j$  Why are Vandermonde-matrices with ☀ always invertible?  I have tried to find a short argument for that. I know some ways to show that in principle:   rank is equal to dimension all lines / rows are linear independence determinant is not zero find inverse   According to , the determinant is  $$\displaystyle V_n = \prod_{1 \le i &lt; j \le n} \left({x_j - x_i}\right)$$  There are two proofs for this determinant, but I've wondered if there is a simpler way to show that such matrices are invertible.
5663
How do I prove that $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$? $\mathbb{Q}$ is the rational field.  I want to know the detail about the proof. Thanks in advance.  Actually I know any two of them and three of them are linearly independent.
113065
I know that any subset of V with more than k vectors is always linearly dependent, but how do I include matrix A into the vectors given to prove the statement?   Also, for part (b), does the statement mean that Ax=0 only has the trivial solution since V intersects span{u1, u2, ..., uk}={0}?  Any help would be appreciated! Thank you!
1470
This question aims to create an &quot;&quot;  of numerous questions that ask about determinants of specific matrices (I may have missed a few):              The general question of this type is  Let $A$ be a square matrix of rank$~1$, let $I$ the identity matrix of the same size, and $\lambda$ a scalar. What is the determinant of $A+\lambda I$?  A clearly very closely related question is  What is the characteristic polynomial of a matrix $A$ of rank$~1$?
93380
Say I have a matrix A that is m by n and m &lt; n. For context, say I am interested in finding a vector $ \theta_z \in \mathbb{R}^m$ such that:  $$ A^T \theta_z = \theta_x$$  i.e. we have more rows/"equations" than columns/"unknowns".  I was going to do it by multiplying both sides by A and then find the inverse of $AA^T$.  I was told that this only works when we have linearly independent rows. Is this suppose to be obvious? What is an intuitive way to explain this? Also, I am even more interested in a precise and rigorous proof if there is one.   I was also told that when the rows are dependent, then  the system has an infinite number of solutions. However, I was a little skeptical about this, because if we have more rows, then if we did some elimination on the original $ A^T \theta_z = \theta_x$, then we could have at most a rank of size $m$ (one for each column). Which could lead to only 1 solution or zero (none) but never to infinity. So I think that solutions is wrong, i.e. linearly independent rows if anything, shouldn't lead to more than 2 solutions ever, we should only have 1 or zero for the original system $ A^T \theta_z = \theta_x$. Is that correct?
76871
I am trying to prove that $a_1$, $a_2$, $a_3$ are linearly independent.  I am asked to use vector product and prove that if $c_{1}a_{1} + c_{2}a_{2} + c_{3}a_{3} = 0$ then $c_1 = c_2 = c_3 = 0$  I am completely stuck on where to go with this problem. I would think that linearly independent then the null space of the space can only be equal to $0$. Would this prove it?
49485
Let $u_1, u_2,... , u_k$ be linearly independent vectors in $\Bbb R^n$.  Let $A$ be an $m × n$ matrix and $V$ the solution space of the homogeneous linear system $Ax = 0$  Prove that if$V \cap \text{Span}\{u_1, u_2,...,u_k\} = \{ 0\} $, then $Au_1, Au_2,...,Au_k$ are linearly independent.  I have an idea that to prove linearly independence of $Au_1, Au_2, ..., Au_k$ I have to show that the equation $c_1Au_1, c_2Au_2, ..., c_kAu_k = 0, c_1, c_2,...,c_k \in \Bbb R$ , only has the trivial solution $c_1=c_2=...=c_k=0$  any help is appreciated!
95009
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 is square) would be invertible and would have a non-zero determinant. However, what confuses me is, what sort of conditions must we have on each row of $X$ such that $A$ would be invertible.   It would be very nice to have a solution of the form:   when $n &gt; m$ then $X$ must have... when $n &lt; m$ then $X$ must have... when $n = m$ then $X$ must have...   I think in the 3rd case we just need $X$ to be invertible but I was unsure of the other two cases.