Matrix Algebra and Definitions: Learning Objectives 5, 6
Question 1: Let \[A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, B = \begin{bmatrix} 1 & 2\\3&4\\5&6\\7&8 \end{bmatrix}, C= \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8 \end{bmatrix} \]
Find the transpose of each matrix.
The transpose of a matrix is found by interchanging its rows and columns. The first row becomes the first column, the second row becomes the second column, and so on. An $m \times n$ matrix becomes an $n \times m$ matrix.
a) $A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$
b) $B^T = \begin{bmatrix} 1 & 3 & 5 & 7\\ 2 & 4 & 6 & 8 \end{bmatrix}$
c) $C^T = \begin{bmatrix} 1 & 2 \\ 4 & 5 \\ 7 & 8 \end{bmatrix}$
b) $B^T = \begin{bmatrix} 1 & 3 & 5 & 7\\ 2 & 4 & 6 & 8 \end{bmatrix}$
c) $C^T = \begin{bmatrix} 1 & 2 \\ 4 & 5 \\ 7 & 8 \end{bmatrix}$
Question 2: Let \[ A = \begin{bmatrix} -4 & 9 & 4\\ -3 & 10 & -4 \\ 1 & 0 & 6 \end{bmatrix}, B= \begin{bmatrix} -3 &-8 & 10\\ 1 & -5 & -9 \\ -10 & 2 & 4 \end{bmatrix}, C= \begin{bmatrix} -1 & 5 & 4\\ 8 & 1 & 9 \\ -6 & 8 & -3 \end{bmatrix}\]
Evaluate the following:
- $(A+B)^T$
- $(AB)^T$
- $(C^TA)^T$
You can either perform the operations inside the parentheses first and then take the transpose, or use the properties of transposes, such as $(X+Y)^T = X^T+Y^T$ and $(XY)^T = Y^TX^T$.
- $\begin{bmatrix} -7 & -2 & -9\\ 1 & 5 & 2 \\ 14 & -13 & 10 \end{bmatrix}$
- $\begin{bmatrix} -19 & 59 & -63\\ -5 & -34 & 4 \\ -105 & -136 & 34 \end{bmatrix}$
- $\begin{bmatrix} -26 & 71 & -72\\ -15 & 55 & 64 \\ -46 & 126 & -38 \end{bmatrix}$
a) $A+B = \begin{bmatrix} -7 & 1 & 14\\ -2 & 5 & -13 \\ -9 & 2 & 10 \end{bmatrix}$. So, $(A+B)^T = \begin{bmatrix} -7 & -2 & -9\\ 1 & 5 & 2 \\ 14 & -13 & 10 \end{bmatrix}$.
b) $AB = \begin{bmatrix} -19 & -5 & -105 \\ 59 & -34 & -136 \\ -63 & 4 & 34 \end{bmatrix}$. So, $(AB)^T = \begin{bmatrix} -19 & 59 & -63\\ -5 & -34 & 4 \\ -105 & -136 & 34 \end{bmatrix}$.
c) Using the property $(XY)^T = Y^TX^T$, we have $(C^TA)^T = A^T(C^T)^T = A^TC$. \[ A^TC = \begin{bmatrix} -4 & -3 & 1\\ 9 & 10 & 0 \\ 4 & -4 & 6 \end{bmatrix} \begin{bmatrix} -1 & 5 & 4\\ 8 & 1 & 9 \\ -6 & 8 & -3 \end{bmatrix} = \begin{bmatrix} 4-24-6 & -20-3+8 & -16-27-3 \\ -9+80 & 45+10 & 36+90 \\ -4-32-36 & 20-4+48 & 16-36-18 \end{bmatrix} = \begin{bmatrix} -26 & -15 & -46\\ 71 & 55 & 126 \\ -72 & 64 & -38 \end{bmatrix} \]
b) $AB = \begin{bmatrix} -19 & -5 & -105 \\ 59 & -34 & -136 \\ -63 & 4 & 34 \end{bmatrix}$. So, $(AB)^T = \begin{bmatrix} -19 & 59 & -63\\ -5 & -34 & 4 \\ -105 & -136 & 34 \end{bmatrix}$.
c) Using the property $(XY)^T = Y^TX^T$, we have $(C^TA)^T = A^T(C^T)^T = A^TC$. \[ A^TC = \begin{bmatrix} -4 & -3 & 1\\ 9 & 10 & 0 \\ 4 & -4 & 6 \end{bmatrix} \begin{bmatrix} -1 & 5 & 4\\ 8 & 1 & 9 \\ -6 & 8 & -3 \end{bmatrix} = \begin{bmatrix} 4-24-6 & -20-3+8 & -16-27-3 \\ -9+80 & 45+10 & 36+90 \\ -4-32-36 & 20-4+48 & 16-36-18 \end{bmatrix} = \begin{bmatrix} -26 & -15 & -46\\ 71 & 55 & 126 \\ -72 & 64 & -38 \end{bmatrix} \]
Question 3: Let $X$ and $Y$ be $n\times n$ matrices. For each of the following equalities, show that the given property is true or provide a counterexample.
- $(X+Y)^T = X^T +Y^T$
- $(XY)^T = X^TY^T$
- $(XY)^T = Y^TX^T$
- $(cX)^T = c(X^T)$
- $(cX)^T = \frac{1}{c}X^T$
You can prove the true properties using an element-wise argument. For the false ones, try to construct a simple $2 \times 2$ counterexample.
- True.
- False.
- True.
- True.
- False.
a) True. The transpose of a sum is the sum of the transposes. $[(X+Y)^T]_{ij} = (X+Y)_{ji} = X_{ji}+Y_{ji} = (X^T)_{ij}+(Y^T)_{ij} = [X^T+Y^T]_{ij}$.
b) False. The order of multiplication is reversed. Let $X = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ and $Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$. Then $(XY)^T = \begin{bmatrix} 4 & 1 \\ 2 & 1 \end{bmatrix}$, but $X^TY^T = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}$.
c) True. This is the correct property for the transpose of a product. Let $Z = XY$. Then $Z_{ij} = \sum_k X_{ik}Y_{kj}$. The element in row $j$, column $i$ of $Z^T$ is $Z_{ji} = \sum_k X_{jk}Y_{ki}$. Now consider $Y^TX^T$. The element in row $j$, column $i$ is $\sum_k (Y^T)_{jk}(X^T)_{ki} = \sum_k Y_{kj}X_{ik}$, which is the same.
d) True. A scalar can be factored out of the transpose operation. $[(cX)^T]_{ij} = (cX)_{ji} = c \cdot X_{ji} = c \cdot (X^T)_{ij} = [c(X^T)]_{ij}$.
e) False. Let $X$ be the $2 \times 2$ identity matrix ($I_2$) and $c = 2$. Since $X$ is symmetric $X^T = X$. However $(cX)^T = 2 I_2 \neq \frac{1}{2} I_2 = \frac{1}{c}X^T$.
b) False. The order of multiplication is reversed. Let $X = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ and $Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$. Then $(XY)^T = \begin{bmatrix} 4 & 1 \\ 2 & 1 \end{bmatrix}$, but $X^TY^T = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}$.
c) True. This is the correct property for the transpose of a product. Let $Z = XY$. Then $Z_{ij} = \sum_k X_{ik}Y_{kj}$. The element in row $j$, column $i$ of $Z^T$ is $Z_{ji} = \sum_k X_{jk}Y_{ki}$. Now consider $Y^TX^T$. The element in row $j$, column $i$ is $\sum_k (Y^T)_{jk}(X^T)_{ki} = \sum_k Y_{kj}X_{ik}$, which is the same.
d) True. A scalar can be factored out of the transpose operation. $[(cX)^T]_{ij} = (cX)_{ji} = c \cdot X_{ji} = c \cdot (X^T)_{ij} = [c(X^T)]_{ij}$.
e) False. Let $X$ be the $2 \times 2$ identity matrix ($I_2$) and $c = 2$. Since $X$ is symmetric $X^T = X$. However $(cX)^T = 2 I_2 \neq \frac{1}{2} I_2 = \frac{1}{c}X^T$.
Question 4: Let \( M = \begin{bmatrix} 3 & -1 & 2\\ 0 & 4 & 5 \\ -2 & 1 & 6 \end{bmatrix} \). The trace of a square matrix is defined as the sum of the elements on its main diagonal.
- Use this definition to compute $tr(M)$, the trace of the matrix $M$.
- Let $N = 2M$, compute tr($N$).
- Show that in general $tr(kM)=k\cdot tr(M)$ for any scalar $k$.
- Without computing $M^T$, explain why tr($M^T$) = tr($M$).
The main diagonal runs from the top-left to the bottom-right. The linearity property of trace means $tr(cA) = c \cdot tr(A)$. The transpose operation swaps off-diagonal elements.
- 13
- 26
- The proof follows from the definitions of scalar multiplication and trace.
- The diagonal elements remain unchanged by the transpose operation.
a) The elements on the main diagonal are 3, 4, and 6.
\[ tr(M) = 3 + 4 + 6 = 13 \]
b) $N = 2M = \begin{bmatrix} 6 & -2 & 4\\ 0 & 8 & 10 \\ -4 & 2 & 12 \end{bmatrix}$.
\[ tr(N) = 6 + 8 + 12 = 26 \]
c) Let $M$ be an $n \times n$ matrix.
\[ tr(kM) = \sum_{i=1}^n (kM)_{ii} = \sum_{i=1}^n k \cdot M_{ii} = k \sum_{i=1}^n M_{ii} = k \cdot tr(M) \]
d) The transpose of a matrix swaps elements across the main diagonal (i.e., it swaps $m_{ij}$ and $m_{ji}$). The elements on the main diagonal ($m_{ii}$) are not affected by this operation. Since the trace is the sum of only the diagonal elements, and these elements are unchanged, the trace remains the same.
Question 5: Let $X$ and $Y$ be $n \times n$ matrices.
- Show that tr($X+Y$) = tr($Y+X$).
- Show that tr($XY$) = tr($YX$).
For part (a), use the property that $tr(A+B)=tr(A)+tr(B)$ and the commutativity of scalar addition. For part (b), write out the formula for a diagonal element of a matrix product, $(AB)_{ii}$, using summation notation. Then write out the formula for the trace and see if you can manipulate the summations.
a) We use two properties of the trace: $tr(A+B) = tr(A)+tr(B)$ and the fact that the trace is a scalar.
\[ tr(X+Y) = tr(X) + tr(Y) \]
Since addition of real numbers is commutative, $tr(X)+tr(Y) = tr(Y)+tr(X)$.
\[ tr(Y)+tr(X) = tr(Y+X) \]
Therefore, $tr(X+Y) = tr(Y+X)$.
b) The element on the diagonal of the product $XY$ is given by: \[ (XY)_{ii} = \sum_{k=1}^n x_{ik}y_{ki} \] The trace of $XY$ is the sum of these diagonal elements: \[ tr(XY) = \sum_{i=1}^n (XY)_{ii} = \sum_{i=1}^n \sum_{k=1}^n x_{ik}y_{ki} \] Now, let's look at the trace of $YX$. The element on the diagonal of $YX$ is: \[ (YX)_{kk} = \sum_{i=1}^n y_{ki}x_{ik} \] The trace of $YX$ is the sum of these diagonal elements: \[ tr(YX) = \sum_{k=1}^n (YX)_{kk} = \sum_{k=1}^n \sum_{i=1}^n y_{ki}x_{ik} \] Since the order of summation can be interchanged and scalar multiplication is commutative ($y_{ki}x_{ik} = x_{ik}y_{ki}$), the two expressions are equivalent: \[ \sum_{i=1}^n \sum_{k=1}^n x_{ik}y_{ki} = \sum_{k=1}^n \sum_{i=1}^n x_{ik}y_{ki} \] Therefore, $tr(XY) = tr(YX)$.
b) The element on the diagonal of the product $XY$ is given by: \[ (XY)_{ii} = \sum_{k=1}^n x_{ik}y_{ki} \] The trace of $XY$ is the sum of these diagonal elements: \[ tr(XY) = \sum_{i=1}^n (XY)_{ii} = \sum_{i=1}^n \sum_{k=1}^n x_{ik}y_{ki} \] Now, let's look at the trace of $YX$. The element on the diagonal of $YX$ is: \[ (YX)_{kk} = \sum_{i=1}^n y_{ki}x_{ik} \] The trace of $YX$ is the sum of these diagonal elements: \[ tr(YX) = \sum_{k=1}^n (YX)_{kk} = \sum_{k=1}^n \sum_{i=1}^n y_{ki}x_{ik} \] Since the order of summation can be interchanged and scalar multiplication is commutative ($y_{ki}x_{ik} = x_{ik}y_{ki}$), the two expressions are equivalent: \[ \sum_{i=1}^n \sum_{k=1}^n x_{ik}y_{ki} = \sum_{k=1}^n \sum_{i=1}^n x_{ik}y_{ki} \] Therefore, $tr(XY) = tr(YX)$.