1 Matrices and Basic Operations

“It is not enough to have a good mind; the main thing is to use it well.”
– René Descartes

Handout

1.1 Important Properties

Addition/Subtraction

Commutativity: A\pm B=B\pm A
Associativity: (A+B)+C=A+(B+C)
Identity: A+0=0+A=A

Matrix Multiplication

Non-Commutativity: AB\neq BA, in general
Associativity: (AB)C=A(BC)
Identity: AI=IA=A
Zero: A0=0A=0
Distribution: A(B+C)=AB+AC and (A+B)C=AC+BC

Transpose

Associativity: k(A+B)=kA+kB
Zero: 0A=0

Multiplication by Scalar

Idempotent: (A^T)^T=A
Distribution over Sum: (A+B)^T=A^T+B^T
Distribution over Product: (AB)^T=B^TA^T

Inverse (for invertible matrices)

Idempotent: (A^{-1})^{-1}=A
Distribution over Product: (AB)^{-1}=B^{-1}A^{-1}
Transpose: (A^{-1})^T=(A^T)^{-1}

Determinant

Distribution over Product: det(AB)=det(A) det(B)

1.2 Exercises

Exercise 1.1 Can two matrices be always added?

Solution. No, the matrices must have the same size, i.e., the same number of rows and columns.

Exercise 1.2 If addition is possible, add the following matrices: A=\begin{bmatrix} -1 & 2 & 3\\ 0 & 1 & 5 \end{bmatrix} \text{ and } B=\begin{bmatrix} 1/2 & 1 & 3\\ 2 & -2 & 0 \end{bmatrix}.

Solution. The matrix addition is possible since both the matrices have 2 rows and 3 columns.

A+B=\begin{bmatrix} -1 + 1/2 & 2 + 1 & 3 + 3\\ 0 + 2 & 1 + (-2) & 5 + 0 \end{bmatrix} = \begin{bmatrix} -1/2 & 3 & 6\\ 2 & -1 & 5 \end{bmatrix}.

Exercise 1.3 For matrices A,B, is it always true that A+B=B+A? Explain your answer.

Exercise 1.4 If multiplication is possible, multiply the following matrices: A=\begin{bmatrix} 1 & 2 & 3\\ 0 & -1 & 5 \end{bmatrix} \text{ and } B=\begin{bmatrix} 1 & 2 & 1\\ 1 & 0 & 2 \end{bmatrix}.

Exercise 1.5 If multiplication is possible, multiply the following matrices: A=\begin{bmatrix} 1 & 2\\ 0 & -1\\ -1 & 3 \end{bmatrix} \text{ and } B=\begin{bmatrix} 1 & 5 & 1\\ 1 & 12 & 2 \end{bmatrix}.

Exercise 1.6 For matrices A and B, is it always true that AB=BA? Explain your answer.

Exercise 1.7 Give an example of a 4\times 2 diagonal matrix.

Exercise 1.8 Intuitively justify the following fact: (AB)^T=B^TA^T.

Exercise 1.9 Is the sum of two symmetric matrices always symmetric?

Exercise 1.10 Is the product of two symmetric matrices always symmetric?

Solution. The answer is no. For symmetric matrices A and B, the product is symmetric iff AB=BA. (Hint: (AB)^T=B^TA^T)

Exercise 1.11 Give an example of a skew-symmetric matrix of order 4\times 4 with at least 4 non-zero elements.

Solution. (Hint: What does the diagonal of a skew symmetric matrix look like?)

Exercise 1.12 For any matrix A, explain why

A+A^T is symmetric.
A-A^T is skew-symmetric.

Solution.

Putting B=A+A^T, we have B^T=(A+A^T)^T=A^T+(A^T)^T = A^T+A=B
Similar

Exercise 1.13 If A has an inverse, then show that the inverse is unique.

Exercise 1.14 If A is the inverse of B, then show that B is the inverse of A.

Exercise 1.15

Is \text{det} (A+B)=\text{det} (A)+ \text{det} (B)?
Is \text{det} (AB)=\text{det} (A) \text{det} (B)?

Solution.

the 2 \times 2 case can be checked directly. How do you prove the n \times n case?

Exercise 1.16 Suppose A is a 2 \times 2 matrix and B is the matrix we get by interchanging the two rows of A. Show that \text{det} B = - \text{det} A.

Exercise 1.17 Find the determinant of \begin{bmatrix}-1 & 2 & 3\\ 0 & 1 & 5\\ 4 & 1 & -1\end{bmatrix}.

Exercise 1.18 Find a 3\times 3 matrix whose determinant is 6 and the absolute value of each entry is no bigger than 3.

Solution. (Hint: what is the determinant of a (upper) triangular matrix?)

Exercise 1.19 \begin{bmatrix} 2 & 1 & 3 \end{bmatrix}\begin{bmatrix} -1 & 0 & -1\\ -1 & 1 & 0\\ 0 & 1 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}=?

Exercise 1.20 Find the inverse of the following matrix: A=\begin{bmatrix} 1 & 2 & 3\\ 0 & -1 & 2\\ 1 & -2 & 0 \end{bmatrix}.

Exercise 1.21 Find the inverse of the following matrix: \begin{bmatrix} -1 & 0 & -1\\ -1 & 1 & 0\\ 0 & 1 & 1 \end{bmatrix}.

Exercise 1.22 If =\begin{bmatrix} 5 & 3\\ -1 & -2\\ \end{bmatrix}, then show that A satisfies the equation A^2-3A-7I=0.

Solution. (Hint: try to factor the LHS)

Exercise 1.23 Consider a n \times n matrix A. Then:

If A is invertible and AB=0 for some n \times n matrix B, then B=0.
If A is not invertible, then we can find a n \times n matrix B with AB=0 but B \ne 0.

Solution.

For any non-invertible 2 \times 2 matrix A= \begin{bmatrix} a & b\\ c & d \end{bmatrix}, where a,b,c,d are not all 0, you can choose B= \begin{bmatrix} -b & -d\\ a & c \end{bmatrix}. How do you do the general case?