Section ROD  Rank One Decomposition

From A First Course in Linear Algebra
Version 1.08
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/

This Section is a Draft, Subject to Changes

Our first decomposition applies only to diagonalizable (Definition DZM) matrices, and yields a decomposition into a sum of very simple matrices.

Theorem ROD
Rank One Decomposition
Suppose that $A$ is a diagonalizable matrix of size $n$ and rank $r$. Then there are $r$ square matrices $A_1,A_2,A_3,\dots ,A_r$, each of size $n$ and rank $1$ such that

Furthermore, if ${\lambda }_1,{\lambda }_2,{\lambda }_3,\dots ,{\lambda }_r$ are the nonzero eigenvalues of $A$, then there are two sets of $r$ linearly independent vectors from ${ℂ}^n$,

such that $A_k={\lambda }_k{x}_k{y}_k^t$, $1\le k\le r$. $\square$

Proof   The proof is constructive. Generally, we will diagonalize $A$, creating a nonsingular matrix $S$ and a diagonal matrix $D$. Then we split up the diagonal matrix into a sum of matrices with a single nonzero entry (on the diagonal). This fundamentally creates the decomposition in the statement of the theorem, the remainder is just bookkeeping. The vectors in $X$ and $Y$ will result from the columns of $S$ and the rows of $S^{-1}$.

Let ${\lambda }_1,{\lambda }_2,{\lambda }_3,\dots ,{\lambda }_n$ be the eigenvalues of $A$ (repeated according to their algebraic multiplicity). If $A$ has rank $r$, then $dim\left(\mathcal{N}\left(A\right)\right)=n-r$ (Theorem RPNC). The null space of $A$ is the eigenspace of the eigenvalue $\lambda =0$ (Theorem EMNS), so it follows that the algebraic multiplicity of $\lambda =0$ is $n-r$, ${\alpha }_A\left(0\right)=n-r$. Presume that the complete list of eigenvalues is ordered so that ${\lambda }_k=0$ for $r+1\le k\le n$.

Since $A$ is hypothesized to be diagonalizable, there exists a diagonal matrix $D$ and an invertible matrix $S$, such that $D=S^{-1}AS$. We can rearrange tis equation to read, $A=SD{S}^{-1}$. Also, the proof of Theorem DC says that the diagonal elements of $D$ are the eigenvalues of $A$ and we have the flexibility to assume they lie on the diagonal in the same order as we have specified above. Now, let $X^{\ast }=\left\{x_1,x_2,x_3,\dots ,x_n\right\}$ be the columns of $S$, and let $Y^{\ast }=\left\{y_1,y_2,y_3,\dots ,y_n\right\}$ be the rows of $S^{-1}$ converted to column vectors. With little motivation other than the statement of the theorem, define size $n$ matrices $A_k$, $1\le k\le n$ by $A_k={\lambda }_k{x}_k{y}_k^t$. Finally, let $D_k$ be the size $n$ matrix that is totally zero, other than having ${\lambda }_k$ in row $k$ and column $k$.

With everything in place, we compute entry-by-entry,

So by Definition ME we have the desired equality of matrices. The careful reader will have noted that $A_k=\mathcal{O}$, $r+1\le k\le n$, since ${\lambda }_k=0$ in these instances. To get the sets $X$ and $Y$ from $X^{\ast }$ and $Y^{\ast }$, simply discard the last $n-r$ vectors. We can safely ignore (or remove) $A_{r+1},A_{r+2},\dots ,A_n$ from the summation just derived.

One last assertion to check. What is the rank of $A_k$, $1\le k\le r$? Every row of $A_k$ is a scalar multiple of $y_k^t$, row $k$ of the nonsingular matrix $S^{-1}$ (Theorem MIMI). As a row of a nonsingular matrix, $y_k^t$ cannot be all zeros. In particular, row $i$ of $A_k$ is obtained as a scalar multiple of $y_k^t$ by the scalar ${\alpha }_k{\left[x_k\right]}_i$. We have restricted ourselves to the nonzero eigenvalues of $A$, and as $S$ is nonsingular, some entry of $x_k$ is nonzero. This all implies that some row of $A_k$ will be nonzero. Now consider row-reducing $A_k$. Swap the nonzero row up into row 1. Use scalar multiples of this row to zero out every other row. This leaves a single nonzero row in the reduced row-echelon form, so $A_k$ has rank one. $■$

We record two observations that was not stated in our theorem above. First, the vectors in $X$, chosen as columns of $S$, are eigenvectors of $A$. Second, the product of two vectors from $X$ and $Y$ in the opposite order, by which we mean $y_i^t{x}_j$, is the entry in row $i$ and column $j$ of the matrix product $S^{-1}S=I_n$ (Theorem EMP). In particular,

We give two computational examples. One small, one a bit bigger.

Example ROD2
Rank one decomposition, size 2
Consider the $2\times 2$ matrix,

By the techniques of Chapter E we find the eigenvalues and eigenspaces,

With $n=2$ distinct eigenvalues, Theorem DED tells us that $A$ is diagonalizable, and with no zero eigenvalues we see that $A$ has full rank. Theorem DC says we can construct the nonsingular matrix $S$ with eigenvectors of $A$ as columns, so we have

From these matrices we obtain the sets of vectors

And we have the matrices,

And you can easily verify that $A=A_1+A_2$. $⊠$

Here’s a slightly larger example, and the matrix does not have full rank.

Example ROD4
Rank one decomposition, size 4
Consider the $4\times 4$ matrix,

By the techniques of Chapter E we find the eigenvalues and eigenvectors,

The algebraic and geometric multiplicities of each eigenvalue are equal, so Theorem DMFE tells us that $A$ is diagonalizable. With a single zero eigenvalue we see that $A$ has rank $4-1=3$. Theorem DC says we can construct the nonsingular matrix $S$ with eigenvectors of $A$ as columns, so we have

Since $r=3$, we need only collect three vectors from each of these matrices,

And we obtain the matrices,

Then we verify that