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}
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}
To determine the rank of the sum and product of matrices, we need to understand the concept of matrix rank.
The rank of a matrix is the maximum number of linearly independent rows or columns in the matrix. In other words, it is the dimension of the vector space spanned by the rows or columns of the matrix.
For the given matrices:
- \mathbf{A} = \begin{pmatrix} -1 & 1 \\ -3 & 3\end{pmatrix}
- \mathbf{B} = \begin{pmatrix} 1 & -1 \\ -1 & 1\end{pmatrix}
- \mathbf{C} = \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}
- \mathbf{D} = \begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}
We can express each of these matrices as outer products as follows:
- \mathbf{A} = \mathbf{u}\mathbf{v}^T
- \mathbf{B} = \mathbf{v}\mathbf{v}^T
- \mathbf{C} = \mathbf{w}\mathbf{w}^T
- \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}
Now, let's consider the combinations of these matrices and determine their ranks.
1. \mathbf{A} + \mathbf{A}
Since \mathbf{A} is a rank-1 matrix, adding it to itself does not change its rank. Thus, \mathbf{A} + \mathbf{A} is also a rank-1 matrix.
2. \mathbf{A} + \mathbf{B}
The sum of two rank-1 matrices could potentially have a higher rank. Let's calculate the sum:
\mathbf{A} + \mathbf{B} = \mathbf{u}\mathbf{v}^T + \mathbf{v}\mathbf{v}^T
= \mathbf{u}\mathbf{v}^T + \mathbf{v}\mathbf{v}^T + \mathbf{0}
= (\mathbf{u} + \mathbf{v})(\mathbf{v}^T)
The rank of the resulting matrix depends on whether \mathbf{u} + \mathbf{v} is linearly independent from \mathbf{v}. In this case, \mathbf{u} + \mathbf{v} = \begin{pmatrix} 0 \\ 4 \end{pmatrix} is linearly independent from \mathbf{v}. Therefore, the rank of \mathbf{A} + \mathbf{B} is 2.
3. \mathbf{A} + \mathbf{C}
Similar to the previous case, let's calculate the sum:
\mathbf{A} + \mathbf{C} = \mathbf{u}\mathbf{v}^T + \mathbf{w}\mathbf{w}^T
= \mathbf{u}\mathbf{v}^T + \mathbf{w}\mathbf{w}^T + \mathbf{0}
= (\mathbf{u} + \mathbf{w})(\mathbf{w}^T)
Again, the rank depends on whether \mathbf{u} + \mathbf{w} is linearly independent from \mathbf{w}. In this case, \mathbf{u} + \mathbf{w} = \begin{pmatrix} 1 \\ 4 \end{pmatrix} is linearly independent from \mathbf{w}. Therefore, the rank of \mathbf{A} + \mathbf{C} is 2.
4. \mathbf{A}\mathbf{B}
The product of two rank-1 matrices does not always result in a rank-2 matrix. In this case:
\mathbf{A}\mathbf{B} = \mathbf{u}\mathbf{v}^T \mathbf{v}\mathbf{v}^T
= \mathbf{u}(\mathbf{v}^T \mathbf{v})\mathbf{v}^T
= (\mathbf{u} \cdot \mathbf{v})\mathbf{v}\mathbf{v}^T
= (\mathbf{u} \cdot \mathbf{v})\mathbf{v}\mathbf{v}^T + \mathbf{0}
= (\mathbf{u} \cdot \mathbf{v})(\mathbf{v}\mathbf{v}^T)
Since \mathbf{u} \cdot \mathbf{v} = -1 and \mathbf{v}\mathbf{v}^T is a rank-1 matrix, the resulting matrix is also rank 1.
5. \mathbf{A}\mathbf{C}
Following similar steps as above, we have:
\mathbf{A}\mathbf{C} = (\mathbf{u} \cdot \mathbf{w})(\mathbf{w}\mathbf{w}^T)
Since \mathbf{u} \cdot \mathbf{w} = 0 and \mathbf{w}\mathbf{w}^T is a rank-1 matrix, the resulting matrix is also rank 1.
6. \mathbf{B}\mathbf{D}
The product of two rank-1 matrices could potentially have a higher rank. Let's calculate the product:
\mathbf{B}\mathbf{D} = \mathbf{v}\mathbf{v}^T \mathbf{x}\mathbf{x}^T
= \mathbf{v}(\mathbf{v}^T \mathbf{x})\mathbf{x}^T
= (\mathbf{v} \cdot \mathbf{x})\mathbf{v}\mathbf{x}^T
= (\mathbf{v} \cdot \mathbf{x})(\mathbf{v}\mathbf{x}^T)
Since \mathbf{v} \cdot \mathbf{x} = 0, the resulting matrix is a rank-1 matrix.
In summary:
The combination of matrices that have rank 2 is:
- \mathbf{A} + \mathbf{B}
The combinations of matrices that have rank 1 are:
- \mathbf{A} + \mathbf{A}
- \mathbf{A} + \mathbf{C}
- \mathbf{A}\mathbf{B}
- \mathbf{A}\mathbf{C}
- \mathbf{B}\mathbf{D}