This question is meant to serve as an answer to the following: If you sum two rank-1 matrices, do you get a rank-2 matrix? What about products? More generally, what rank is the sum of a rank-r_1 and a rank-r_2 matrix?"

Let \mathbf{A}= \begin{pmatrix} -1 & 1 \\ -3 & 3\end{pmatrix}, \mathbf{B}= \begin{pmatrix} 1 & -1 \\ -1 & 1\end{pmatrix} and \mathbf{C}= \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix} and \mathbf{D}= \begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}. Observe that all four of these matrices are rank 1.

There are many ways to determine rank. Here is one useful fact that you could use for this problem:

“Every rank-1 matrix can be written as an outer product. Conversely, every outer product \mathbf{u}\mathbf{v}^ T is a rank-1 matrix."
For example, \mathbf{A}= \mathbf{u}\mathbf{v}^ T, \mathbf{B}= \mathbf{v}\mathbf{v}^ T, \mathbf{C}= \mathbf{w}\mathbf{w}^ T and \mathbf{D}= \mathbf{x}\mathbf{x}^ T, where

\mathbf{u}= \begin{pmatrix} 1 \\ 3 \end{pmatrix}, \mathbf{v}= \begin{pmatrix} -1 \\ 1 \end{pmatrix}, \mathbf{w}= \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \mathbf{x}= \begin{pmatrix} 1 \\ 1 \end{pmatrix}.

Which combination of these matrices has rank 2? Choose all that apply.

\mathbf{A}+\mathbf{A}

\mathbf{A}+\mathbf{B}

\mathbf{A}+\mathbf{C}

\mathbf{A}\mathbf{B}

\mathbf{A}\mathbf{C}

\mathbf{B}\mathbf{D}
incorrect
Which combination of these matrices has rank 1? Choose all that apply.

\mathbf{A}+\mathbf{A}

\mathbf{A}+\mathbf{B}

\mathbf{A}+\mathbf{C}

\mathbf{A}\mathbf{B}

\mathbf{A}\mathbf{C}

\mathbf{B}\mathbf{D}

1 answer

For a matrix to have a rank of 2, it must require at least two linearly independent vectors as columns or rows.

The sum of two rank-1 matrices will not necessarily result in a rank-2 matrix. Let's check each combination:

1) \mathbf{A}+\mathbf{A}:
This results in the matrix \begin{pmatrix} -1 & 1 \\ -3 & 3\end{pmatrix} + \begin{pmatrix} -1 & 1 \\ -3 & 3\end{pmatrix} = \begin{pmatrix} -2 & 2 \\ -6 & 6\end{pmatrix}. The rank of this matrix is still 1.

2) \mathbf{A}+\mathbf{B}:
This results in the matrix \begin{pmatrix} -1 & 1 \\ -3 & 3\end{pmatrix} + \begin{pmatrix} 1 & -1 \\ -1 & 1\end{pmatrix} = \begin{pmatrix} 0 & 0 \\ -4 & 4\end{pmatrix}. The rank of this matrix is still 1.

3) \mathbf{A}+\mathbf{C}:
This results in the matrix \begin{pmatrix} -1 & 1 \\ -3 & 3\end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix} -1 & 1 \\ -3 & 4\end{pmatrix}. The rank of this matrix is still 1.

4) \mathbf{A}\mathbf{B}:
This results in the matrix \begin{pmatrix} -1 & 1 \\ -3 & 3\end{pmatrix} \begin{pmatrix} 1 & -1 \\ -1 & 1\end{pmatrix} = \begin{pmatrix} -2 & 2 \\ -6 & 6\end{pmatrix}. The rank of this matrix is still 1.

5) \mathbf{A}\mathbf{C}:
This results in the matrix \begin{pmatrix} -1 & 1 \\ -3 & 3\end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 3\end{pmatrix}. The rank of this matrix is still 1.

6) \mathbf{B}\mathbf{D}:
This results in the matrix \begin{pmatrix} 1 & -1 \\ -1 & 1\end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}. The rank of this matrix is 0.

So, the only combination that results in a rank-2 matrix is:

\mathbf{B}\mathbf{D}
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