Exercise 1
What rows or columns or matrices do you multiply to find
- the second column of $ AB $ ?
- the first row of $ AB $ ?
- the entry in row $ 3 $, column $ 5 $ of $ AB $ ?
- the entry in row $ 1 $, column $ 1 $ of $ CDE $ ?
To find the second column of $ AB $, we multiply matrix $ A $ by the second column of matrix $ B $.
Multiply the first row of $ A $ by matrix $ B $.
Multiply the third row of $ A $ and the fifth column of $ B $.
Multiply first the first row of $ C $ by $ D $, then product the first column of $ E $.
Exercise 2
Show that $ (A + B)^2 $ is different from $ A^2 + 2AB + B^2 $, when $ A=\begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix} $ and $ B=\begin{pmatrix} 1 & 0 \\ 3 & 0 \end{pmatrix} $.
Write down the correct rule for $ (A + B)(A + B) = A^2 + \_\_\_\_\_\_\_\_ + B^2 $.
First, we find the sum of matrices $ A $ and $ B $ : $$ A+B = \begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 3 & 0 \end{pmatrix} = \begin{pmatrix} 1+1 & 2+0 \\ 0+3 & 0+0 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 3 & 0 \end{pmatrix} $$ Next, we compute the square of $ (A+B) $ : $$ (A+B)^2 = \begin{pmatrix} 2 & 2 \\ 3 & 0 \end{pmatrix} \begin{pmatrix} 2 & 2 \\ 3 & 0 \end{pmatrix} = \begin{pmatrix} 10 & 4 \\ 6 & 6 \end{pmatrix} $$ Then we need to compute $ A^2 $, $ B^2 $, and $ 2AB $ separately.
$$ A^2 = \begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix}, B^2 = \begin{pmatrix} 1 & 0 \\ 3 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 3 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 3 & 0 \end{pmatrix} $$ $$ AB = \begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 3 & 0 \end{pmatrix} = \begin{pmatrix} 7 & 0 \\ 0 & 0 \end{pmatrix} $$ Thus, we can calculate the result $$ A^2 + 2AB + B^2 = \begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 14 & 0 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 3 & 0 \end{pmatrix} = \begin{pmatrix} 16 & 2 \\ 3 & 0 \end{pmatrix} $$ We found: $ (A+B)^2 = \begin{pmatrix} 10 & 4 \\ 6 & 6 \end{pmatrix} ,A^2 + 2AB + B^2 = \begin{pmatrix} 16 & 2 \\ 3 & 0 \end{pmatrix} $.
Since $ \begin{pmatrix} 10 & 4 \\ 6 & 6 \end{pmatrix} \neq \begin{pmatrix} 16 & 2 \\ 3 & 0 \end{pmatrix} $, we have shown that $ (A + B)^2 $ is different from $ A^2 + 2AB + B^2 $ for the given matrices. This difference arises because matrix multiplication is generally not commutative ($ AB \neq BA $).
The correct rule for expanding $ (A+B)(A+B) $ for matrices $ A $ and $ B $ is: $ (A + B)(A + B) = A^2 + \underline{AB+BA} + B^2 $.
Exercise 3
If you do a row operation on $ A $ and then a column operation, the result is the same as if you did the column operation first. Why is this true?
Proof:
Performing a row operation on a matrix $ A $ is equivalent to multiplying $ A $ on its left by a corresponding elementary matrix, let’s call it $ E_R $. So, the operation results in $ E_R A $. Then, perform the column operation on the result $ E_R A $. This means multiplying $ E_R A $ on the right by $ E_C $, giving the final result $ (E_R A) E_C $.
Performing a column operation on a matrix $ A $ is equivalent to multiplying $ A $ on its right by a corresponding elementary matrix, let’s call it $ E_C $. So, the operation results in $ A E_C $. Then, perform the row operation on the result $ A E_C $. This means multiplying $ A E_C $ on the left by $ E_R $, giving the final result $ E_R (A E_C) $.
According to the associativity of matrix multiplication, we have $ (E_R A) E_C = E_R (A E_C) $. Therefore, the final matrix result is the same regardless of the order in which the row and column operations are performed.
Exercise 4
Let $ A=\begin{pmatrix} 2 & 3 \\ 1 & 2 \\ 7 & 100 \end{pmatrix} $. Prove that there is no $ 2 \times 3 $ matrix $ B $ such that $ AB = I $.
(Please only use the materials we have learnt so far, in particular the geometric interpretation of $ Ab $ is a linear combination of the column vectors of $ A $).
Proof:
Firstly, if $ AB=I $, $ I $ must be the $ 3 \times 3 $ identity matrix ($ I_3 $).
Considering the column space of matrix $ A $, which has two column vectors: $ A_1 = \begin{pmatrix} 2 \\ 1 \\ 7 \end{pmatrix} $ and $ A_2 = \begin{pmatrix} 3 \\ 2 \\ 100 \end{pmatrix} $. These two vectors are linearly independent, so the column space $ Col(A) $ is a two-dimensional subspace of $ \mathbb{R}^3 $ (a plane through the origin).
If $ AB=I_3 $, then each column of $ I_3 $ must be in the column space of $ A $. That is, for each standard basis vector $ e_j $ (the columns of $ I_3 $), there must exist a column vector $ b_j $ from $ B $ such that $ A b_j = e_j $. This means $ e_j $ must be a linear combination of $ A $’s column vectors, and thus $ e_j \in Col(A) $.
The columns of $ I_3 $ are $ e_1=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} $, $ e_2=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} $, and $ e_3=\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} $. These three vectors are linearly independent and span the entire $ \mathbb{R}^3 $. However, $ Col(A) $ is only a two-dimensional subspace. A two-dimensional subspace cannot contain three linearly independent vectors that span a three-dimensional space.
Therefore, there is no $ 2 \times 3 $ matrix $ B $ such that $ AB=I $.
Exercise 5
Let $ m, n \ge 1 $ and $ A, B $ two $ m \times n $ matrices. Prove that if for all $ x \in \mathbb{R}^n $ we have $ Ax = Bx $, then $ A = B $.
Proof:
We are given $ Ax = Bx $ for all $ x \in \mathbb{R}^n $. This can be rewritten as $ (A-B)x = 0 $ for all $ x \in \mathbb{R}^n $. Let . Then the condition becomes for all .
Consider the standard basis vectors $ e_1, e_2, \ldots, e_n $ in $ \mathbb{R}^n $. Since $ Cx=0 $ for all $ x $, it must hold for each $ e_j $. Therefore, implies that every column of is the zero vector.
If all columns of matrix $ C $ are the zero vector, then $ C $ must be the zero matrix. Since $ C=A-B $, we have $ A-B=0 $, which implies $ A=B $.