2.1 Matrix Operations

Matrix multiplication

Theorem 3

Questions

Let $A=\begin{bmatrix}3 & -6 \\ -1 & 2\end{bmatrix}$, construct a $2\times2$ matrix $B$ such that $AB$ is the zero matrix, where $B$ itself is not the zero matrix.

Using matrix-matrix multiplication properties:

So both vectors are given by $v_i =\begin{bmatrix} x_1\\x_2\end{bmatrix}$ for $Av_i$.

2.2 The Inverse of a Matrix

The inverse $A^{-1}$ of the matrix $A$ is given as:

And we have the following properties:

Theorem 5

If A is an invertible $n\times n$ matrix, then for each $b$ in $R^n$, the equation $Ax=b$ has a unique solution $x=A^{-1}b$

Theorem 7

An invertible matrix $A$ is row equivalent to an identity matrix. Any sequence of elementary row operations that reduces $A$ to $I_n$ also transforms $I_n$ into $A^{-1}$. This means that a matrix must be full-rank to be invertible, which leads into The Invertible Matrix Theorem.

Questions

True or False: A product of two invertible matrices is invertible, and the inverse of the product is the product of their inverses in the same order.

Although it’s true that the product of two invertible matrices is invertible (by Invertible Matrix Theorem), the product is not their inverses in the same order. Remember the actual identity:

Explain why the columns of an $n\times n$ matrix $A$ span $\R^n$ when $A$ is invertible.

By the Invertible Matrix Theorem, invertible matrix $A$ is row equivalent to $I_n$ and is thus full-rank.

Find the inverse of the matrix $A= \begin{bmatrix} 1 & -2 & 1\\ 4 & -7 & 3\\ -2 & 6 & -4\\ \end{bmatrix}$if it exists.

Recall that a matrix must be full rank to be invertible. If we row-reduce this matrix we find that its not invertible.

2.3 Characterizations of Invertible Matrices

Theorem 8: The Invertible Matrix Theorem

For a square matrix $A \in \R^{n\times n}$, either all of the following statements are true or all are false.

1. $A$ is invertible

2. $\det{A} \neq 0$

3. $A$ is row equivalent to $I_n$

4. $A$ is full-rank

5. $Ax=0$ has only the trivial solution

6. $x \mapsto Ax$ is onto AND one-to-one

7. The columns of $A$ span $\R^n$

8. $A^\top$is invertible

Note that a matrix is invertible iff $\det{A} \neq 0$, which is a sufficient condition for invertibility.

Invertible Linear Transformations

If a linear transformation $T(x) = Ax$ is defined by an invertible matrix $A \in \R^{n\times n}$, then the transformation is invertible

where $S(x) = A^{-1}x$.

Note that only square matrices are invertible, thus only $\R^n \rightarrow \R^n$ transformations are invertible.

Questions

Is this matrix invertible?

By Invertible Matrix Theorem, we know that a matrix is invertible iff $\det{A} \neq 0$. We find that this matrix has a determinant of zero, so it’s not invertible.

True or False: If the columns of $A \in \R^{n\times n}$ are linearly independent, then the columns of $A$ span $R^n$.

True. We know by Invertible Matrix Theorem. We also know by Spanning Set Theorem.

True or False: If the linear transformation $x \mapsto Ax$ maps $\R^n$ into $\R^n$ then $A$ has $n$ pivot positions.

True. By Invertible Matrix Theorem, we know that if a linear transformation$\R^n \rightarrow \R^n$ is onto, then the transformation matrix $A$ is an invertible matrix. This gives us the property that it is full-rank, and thus has $n$ pivot positions.

Explain why the columns of $A^2$ span $\R^n$ whenever the columns of $A$ are linearly independent.

By Invertible Matrix Theorem, both of these properties belong to invertible matrices. Using the commutative determinant property:

we know that the product of two invertible matrices is invertible.

Show that the transformation $T$ is invertible and find a formula for $T^{-1}$. $T(x_1,x_2) = (6x_1-8x_2, -5x_1+7x_2)$.

We know that a linear transformation $x \mapsto Ax$ is invertible if $A$ is invertible. Given the definition of $T$, we know the value of $A$:

$A$ is invertible if and only if $\det{A} \neq 0$

$A$ is invertible, thus the linear transformation $T$ defined by the matrix $A$ is invertible.