Find rank of AB, given that A has linearly independent columns and B has rank 2

linear algebramatricesmatrix-rank

I'm trying to prove to myself that given…

Matrix A, which has linearly independent columns, and at least 2 columns…
Matrix B, which has rank of 2

Their product, AB, will have rank of 2. I believe this is because…

Matrix B has two linearly independent columns.
Each column of AB will be a combination of the columns of A
When multiplying matrix A by matrix B, each of the two independent columns of B will create a unique combination of the columns in A.

Is this true? If so, can this be made more rigorous? Thanks for the help in advance!

Best Answer

Yes, it can be made more rigorous/clear, here's how I would do it. Note that it does require thinking about the linear transformations represented by the matrices. I will use abuse notation slightly and use $A$ and $B$ to refer to both the matrices and the linear transformations they represent.

$\newcommand\im{\operatorname{im}}\dim \im B = 2$
$\im AB = A(\im B)$
$A$ is injective (since the columns are linearly independent), so $$\newcommand\rk{\operatorname{rank}}\rk(AB)=\dim \im AB = \dim A(\im B) = \dim\im B = \rk(B) = 2$$

Note that this says more generally that if $A$ has linearly independent columns, then $\rk(AB)=\rk(B)$. Also, I tried to keep the proof as close to what you wrote as possible, to make it clear how it rigorizes what you wrote, but it's probably not the most clear formulation of the proof.

Related Solutions

[Math] Linearly independent rows and matrices

I think your "proof by cases" works just fine, though I personally would do three cases, with first case "if $k < m \implies$ Contradiction", basically, then writing the very reasons you concluded that $k \geq m$.

Note: That's just how I'd do it, since you are already considering cases, we might as well exhaust them all, which you did (consider them all, just not explicitly so).

But my preference has nothing to do with the correctness of your proof. Your logic, which you included, for good reason, does the job.

[Math] How to find linearly independent columns in a matrix

Given $A\in\mathbb{R}^{m\times n}$, $m\geq n$, compute the (economy) QR factorisation. This gives $$ A = QR, \quad R\in\mathbb{R}^{n\times n}. $$ Now if $\mathrm{rank}(A)<n$, the upper triangular matrix $R$ has a staircase profile with some of the "steps" of the staircase over more than one column. Select column indices $j_1,\ldots,j_k$ such that if you remove these columns from $R$, you obtain a nonsingular upper triangular matrix (you can consider it as making each step of the staircase of length 1). The columns $j_1,\ldots,j_k$ can be expressed as linear combination of the remaining columns.

Example: The red columns indicate the columns which are linear combinations of the others.

$$ \begin{bmatrix} \times & \times & \color{red}\times & \times & \color{red}\times & \color{red}\times \\ 0 & \times & \color{red}\times & \times & \color{red}\times & \color{red}\times \\ 0 & 0 & \color{red}0 & \times & \color{red}\times & \color{red}\times \end{bmatrix} $$

Example: For the given matrix from the question, the QR factorisation is:

Q =

         0   -0.4472   -0.8944
         0   -0.8944    0.4472
   -1.0000         0         0

R =

   -1.0000    2.0000   -1.0000
         0    4.4721   -2.2361
         0         0         0

So one can pick the column 2 or 3 to make the matrix $R$ nonsingular and upper triangular (hence either the column 2 or 3 is a linear combination of the others).

Best Answer

Related Solutions

[Math] Linearly independent rows and matrices

[Math] How to find linearly independent columns in a matrix

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