is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Matrix Equation Ax = b


MathJax TeX Test Page If a is an $m \times n$ matrix, with columns $\vec{a_1}, \vec{a_2}, ... , \vec{a_n}$, and $\vec{b}$ is in $\mathbb{R}_m$, $$Ax = b$$ is equivalent to $$\begin{bmatrix}a_1 & a_2 & ... & a_n\end{bmatrix}x$$ which equals $$x_1\vec{a_1} + x_2\vec{a_2} + ... + x_n\vec{a_n} = b$$ So, Ax = b has a solution iff b is a linear combination of the columns of A.

Describing solutions

MathJax TeX Test Page Let $A = \begin{bmatrix}1 & 3 & 4\\-3 & 2& -6\\-3 & -2 & -7\end{bmatrix}$ and $b = \begin{bmatrix}b_1 \\ b_2 \\ b_3\end{bmatrix}$. Describe the conditions when Ax = b. We could create the augmented matrix: $$\begin{bmatrix}1 & 3 & 4 & b_1\\-3 & 2& -6 & b_2\\-3 & -2 & -7 & b_3\end{bmatrix} \rightarrow \begin{bmatrix}1 & 3 & 4 & b_1\\0 & 14& -10 & b_2 + 4b_1\\0 & 0 & 0 & b_3 + 3b_1 - \frac{1}{2}(b_2 + 4b_1)\end{bmatrix}$$ Recall from the last linalg post that a system is consistent iff the last column isn't a pivot column. In other words, if the last row in this matrix ends with a nonzero value, it has no solutions, because it'd be like saying 0 = 1. So we have one condition: $$b_3 + 3b_1 - \frac{1}{2}(b_2 + 4b_1) = b_1 - \frac{1}{2} b_2 + b_3 = 0$$ Now, what if the last row (of the coefficient matrix, so excluding the column of b's) hadn't been all zeroes? What if there was a pivot in every row? Now is where the TFAE comes in.


This stands for The Following Are Equivalent

MathJax TeX Test Page These are either all true or all false. $$\text{For each b in }\mathbb{R}^m \text{, the equation } Ax = b \text{ has a solution.}$$ Remember, the statement above can be false, this just means every other statement is false as well. $$\text{Each b in }\mathbb{R}^m \text{ is a linear combination of the columns of A}$$ This (above) is just a definition, which we showed. $$\text{The columns of A span }\mathbb{R}^m$$ Span is the set of all linear combinations involving A vectors. The span of two vectors (assuming they're linearly independent) is a plane. $$\text{A has a pivot position in every row}$$ This is the only significant one. This explains why the matrix above didn't work for every case. There was no pivot in the third row, so the first statement was false- not every b in $\mathbb{R}^m$ had a solution.


MathJax TeX Test Page There are only two real properties to know. $$A(\vec{u} + \vec{v}) = A\vec{u} + A\vec{v}$$ $$A(c\vec{u}) = c(A\vec{u})$$ These are useful in proofs and understanding other concepts in linear algebra, as we will see in future posts.

LU Factorization

Matrices as Linear Transformations