Matrix Algebra and Definitions: Learning Objectives 3, 4
Question 1: Let \( A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & -7 & 5 \\ -2 & -8 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \).
- Find $AB$ and $BA$.
- $B$ is the $3 \times 3$ identity matrix. What property of identity matrices is being shown in part (a)?
- Let $M \in \mathbb{R}^{m\times n}$, and let $I_n \in \mathbb{R}^{n\times n}$ denote the $n \times n$ identity matrix. Prove that $MI_n = M$.
For part (c), consider the element in the $i$-th row and $j$-th column of the product, $(MI_n)_{ij}$. By definition, this is the dot product of the $i$-th row of $M$ and the $j$-th column of $I_n$.
- $AB = BA = A$.
- This shows that the identity matrix is the multiplicative identity in matrix algebra; multiplying any matrix by it (where defined) leaves the matrix unchanged.
- The proof relies on the definition of matrix multiplication and the structure of the identity matrix.
a) Performing the matrix multiplication:
\[ AB = \begin{bmatrix} 1 & 2 & 3 \\ 2 & -7 & 5 \\ -2 & -8 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & -7 & 5 \\ -2 & -8 & 3 \end{bmatrix} = A \]
Similarly, $BA = A$.
b) This demonstrates the key property of an identity matrix: it serves as the multiplicative identity. For any $n \times n$ matrix $A$, $AI_n = I_nA = A$.
c) Let $C = MI_n$. The element $c_{ij}$ in the $i$-th row and $j$-th column of $C$ is given by the dot product of the $i$-th row of $M$ and the $j$-th column of $I_n$. \[ c_{ij} = \sum_{k=1}^n m_{ik} (I_n)_{kj} \] The identity matrix $I_n$ has entries $(I_n)_{kj} = 1$ if $k=j$ and 0 if $k \neq j$. Therefore, in the summation, all terms are zero except for the one where $k=j$. \[ c_{ij} = m_{i1}(0) + \dots + m_{ij}(1) + \dots + m_{in}(0) = m_{ij} \] Since $c_{ij} = m_{ij}$ for all $i$ and $j$, the matrix $C$ is identical to the matrix $M$. Thus, $MI_n = M$.
b) This demonstrates the key property of an identity matrix: it serves as the multiplicative identity. For any $n \times n$ matrix $A$, $AI_n = I_nA = A$.
c) Let $C = MI_n$. The element $c_{ij}$ in the $i$-th row and $j$-th column of $C$ is given by the dot product of the $i$-th row of $M$ and the $j$-th column of $I_n$. \[ c_{ij} = \sum_{k=1}^n m_{ik} (I_n)_{kj} \] The identity matrix $I_n$ has entries $(I_n)_{kj} = 1$ if $k=j$ and 0 if $k \neq j$. Therefore, in the summation, all terms are zero except for the one where $k=j$. \[ c_{ij} = m_{i1}(0) + \dots + m_{ij}(1) + \dots + m_{in}(0) = m_{ij} \] Since $c_{ij} = m_{ij}$ for all $i$ and $j$, the matrix $C$ is identical to the matrix $M$. Thus, $MI_n = M$.
Question 2: Let \( A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & -7 & 5 \\ -2 & -8 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} \).
- Find $AB$ and $BA$.
- $B$ is a diagonal matrix. What property of diagonal matrices is being demonstrated in part (a)?
Observe how the rows and columns of A are affected when multiplied by the diagonal matrix B on the right versus on the left.
\[ AB = \begin{bmatrix} 1 & 4 & 9 \\ 2 & -14 & 15 \\ -2 & -16 & 9 \end{bmatrix}, \quad BA = \begin{bmatrix} 1 & 2 & 3 \\ 4 & -14 & 10 \\ -6 & -24 & 9 \end{bmatrix} \]
Multiplication by a diagonal matrix on the right scales the columns of the other matrix. Multiplication on the left scales the rows.
a)
\[ AB = \begin{bmatrix} 1 & 2 & 3 \\ 2 & -7 & 5 \\ -2 & -8 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 4 & 9 \\ 2 & -14 & 15 \\ -2 & -16 & 9 \end{bmatrix} \]
\[ BA = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & -7 & 5 \\ -2 & -8 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 4 & -14 & 10 \\ -6 & -24 & 9 \end{bmatrix} \]
b) This demonstrates that multiplying a matrix $A$ by a diagonal matrix $B$ on the right ($AB$) scales the columns of $A$ by the corresponding diagonal entries of $B$. Multiplying on the left ($BA$) scales the rows of $A$ by the corresponding diagonal entries of $B$.
Question 3: Let \( B = \begin{bmatrix} 8 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix} \).
- Find $B^{-1}$.
- What does this show concerning the inverse of a diagonal matrix?
The inverse of a diagonal matrix is found by a simple operation on its diagonal elements.
- \[ B^{-1} = \begin{bmatrix} 1/8 & 0 & 0 \\ 0 & 1/2 & 0 \\ 0 & 0 & 1/4 \end{bmatrix} \]
- The inverse of a diagonal matrix (if it exists) is another diagonal matrix whose diagonal entries are the reciprocals of the original diagonal entries.
Question 4: Let $A = \begin{bmatrix} a & d & e \\ 0 & b & f \\ 0 & 0 & c \end{bmatrix}$ and \( B = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \).
- Compute $AB$ and $BA$.
- $A$ and $B$ are upper triangular matrices. What does part (a) demonstrate about the product of two upper triangular matrices?
- Compute det($A$). What property does this demonstrate about the determinant of an upper triangular matrix?
For part (c), use cofactor expansion along the first column to find the determinant.
- $AB = \begin{bmatrix} a & a+d & a+d+e \\ 0 & b & b+f \\ 0 & 0 & c \end{bmatrix}$, $BA = \begin{bmatrix} a & d+b & e+f+c \\ 0 & b & f+c \\ 0 & 0 & c \end{bmatrix}$
- The product of two upper triangular matrices is also an upper triangular matrix.
- $\det(A) = abc$. The determinant of a triangular matrix is the product of its diagonal entries.
Question 5: Let \( A = \begin{bmatrix} 1 & 1 & -1 \\ 1 & 2 & 0 \\ -1 & 0 & 5 \end{bmatrix} \).
- Find $A^T$.
- $A$ is a symmetric matrix. What property of symmetric matrices is being shown in part (a)?
To find the transpose $A^T$, swap the rows and columns of $A$. Then compare $A^T$ with the original matrix $A$.
- \[ A^T = \begin{bmatrix} 1 & 1 & -1 \\ 1 & 2 & 0 \\ -1 & 0 & 5 \end{bmatrix} \]
- This shows the defining property of a symmetric matrix: it is equal to its own transpose ($A^T = A$).
Question 6: Let \( A = \begin{bmatrix} 4 & -1 \\ 12 & -3 \end{bmatrix} \). Verify whether matrix \( A \) is idempotent, i.e., check whether \( A^2 = A \).
Calculate $A^2$ by multiplying the matrix $A$ by itself.
Yes, the matrix is idempotent.
We check the condition $A^2=A$:
\[ A^2 = A \cdot A = \begin{bmatrix} 4 & -1 \\ 12 & -3 \end{bmatrix} \begin{bmatrix} 4 & -1 \\ 12 & -3 \end{bmatrix} \]
\[ = \begin{bmatrix} 4(4)+(-1)(12) & 4(-1)+(-1)(-3) \\ 12(4)+(-3)(12) & 12(-1)+(-3)(-3) \end{bmatrix} = \begin{bmatrix} 16-12 & -4+3 \\ 48-36 & -12+9 \end{bmatrix} = \begin{bmatrix} 4 & -1 \\ 12 & -3 \end{bmatrix} \]
Since the result is equal to the original matrix $A$, the matrix is idempotent.
Question 7: Let \( P \in \mathbb{R}^{n \times n} \) be an idempotent matrix. Show that \( P(I - P) = \mathbf{0} \), where \( I \) is the \( n \times n \) identity matrix and $\mathbf{0}$ is the zero matrix.
Use the distributive property of matrix multiplication, and then apply the definition of an idempotent matrix ($P^2=P$).
The proof uses the distributive property and the fact that $P^2=P$ for an idempotent matrix.
We start with the expression $P(I-P)$ and use the distributive law for matrices:
\[ P(I - P) = PI - P^2 \]
We know that multiplying any matrix by the identity matrix leaves it unchanged, so $PI = P$.
\[ = P - P^2 \]
By the definition of an idempotent matrix, we know that $P^2=P$. We substitute this into the expression:
\[ = P - P = \mathbf{0} \]
Where $\mathbf{0}$ is the $n \times n$ zero matrix.