Suppose that $M, P$ are two $n \times n$ non-singular matrix. Prove that there is a matrix $N$ such that $MN = P$.

Proof.

As non-singularity and invertibility are equivalent, we know that $M$ has the inverse matrix $M^{-1}$.

Let us think backwards. Suppose that we have $MN=P$ for some matrix $N$, which we want to find.
Then multiply $M^{-1}$ on the left of the equation $MN = P$ yields $N = M^{-1} P$.

This is the matrix we are looking for, as $ M N = M ( M^{-1} P) = P$.

The post If $M, P$ are Nonsingular, then Exists a Matrix $N$ such that $MN=P$ first appeared on Problems in Mathematics.

(a) Let $A$ be a $6\times 6$ matrix and suppose that $A$ can be written as
\[A=BC,\] where $B$ is a $6\times 5$ matrix and $C$ is a $5\times 6$ matrix.

Prove that the matrix $A$ cannot be invertible.

(b) Let $A$ be a $2\times 2$ matrix and suppose that $A$ can be written as
\[A=BC,\] where $B$ is a $ 2\times 3$ matrix and $C$ is a $3\times 2$ matrix.

Can the matrix $A$ be invertible?

Solution.

(a) Prove that the matrix $A$ cannot be invertible.

Since $C$ is a $5 \times 6$ matrix, the equation
\[C\mathbf{x}=\mathbf{0}\] has a nonzero solution $\mathbf{x}_1$.
(There are more variables than equations in the system $C\mathbf{x}=\mathbf{0}$.)

It follows that we have
\begin{align*}
A\mathbf{x}_1=BC\mathbf{x}_1=B\mathbf{0}=\mathbf{0}.
\end{align*}
Since the vector $\mathbf{x}_1$ is nonzero, the matrix $A$ is a singular matrix, hence $A$ is not invertible.

(b) Can the matrix $A$ be invertible?

The answer is yes. For example consider the following $2\times 3$ matrix $B$ and $3 \times 2$ matrix $C$:
\[B=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0
\end{bmatrix}, \qquad C=\begin{bmatrix}
1 & 0 \\
0 & 1 \\
0 &0
\end{bmatrix}.\] Then we have
\begin{align*}
A=BC=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
0 & 1 \\
0 &0
\end{bmatrix}
=
\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix},
\end{align*}
and the matrix $A$ is invertible.

The post If a Matrix is the Product of Two Matrices, is it Invertible? first appeared on Problems in Mathematics.

Suppose that $A$ is a real $n\times n$ matrix.

(a) Is it true that $A$ must commute with its transpose?

(b) Suppose that the columns of $A$ (considered as vectors) form an orthonormal set.
Is it true that the rows of $A$ must also form an orthonormal set?

(University of California, Berkeley, Linear Algebra Qualifying Exam)

Solution.

(a) Is it true that $A$ must commute with its transpose?

The answer is no.

We give a counterexample. Let
\[A=\begin{bmatrix}
1 & -1\\
0& 2
\end{bmatrix}.\] Then the transpose of $A$ is
\[A^{\trans}=\begin{bmatrix}
1 & 0\\
-1& 2
\end{bmatrix}.\] We compute
\[AA^{\trans}=\begin{bmatrix}
1 & -1\\
0& 2
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
-1& 2
\end{bmatrix}
=
\begin{bmatrix}
2 & -2\\
-2& 4
\end{bmatrix},\] and
\[A^{\trans}A=
\begin{bmatrix}
1 & 0\\
-1& 2
\end{bmatrix}
\begin{bmatrix}
1 & -1\\
0& 2
\end{bmatrix}
=
\begin{bmatrix}
1 & -1\\
-1& 5
\end{bmatrix}.
\] Therefore, we see that
\[AA^{\trans}\neq A^{\trans} A,\] that is, $A$ does not commute with its transpose $A^{\trans}$.

(b) Is it true that the rows of $A$ must also form an orthonormal set?

The answer is yes.

Note that in general the column vectors of a matrix $M$ form an orthonormal set if and only if $M^{\trans}M=I$, where $I$ is the identity matrix. (Such a matrix is called orthogonal matrix.)

Thus, by assumption we have $A^{\trans} A=I$. Let $B=A^{\trans}$.
Then the column vectors of $B$ is the row vectors of $A$. Hence it suffices to show that $B^{\trans}B=I$.

Since $A^{\trans} A=I$, we know that $A$ is invertible and the inverse $A^{-1}=A^{\trans}$.
In particular, we have $A^{\trans} A=A A^{\trans}=I$.

We have
\begin{align*}
B^{\trans}B=(A^{\trans})^{\trans}A^{\trans}=(AA^{\trans})^{\trans}=I^{\trans}=I.
\end{align*}
Thus, we obtain $B^{\trans}B=I$ and by the general fact stated above, the column vectors of $B$ form an orthonormal set.
Hence the row column vectors of $A$ form an orthonormal set.

The post If Column Vectors Form Orthonormal set, is Row Vectors Form Orthonormal Set? first appeared on Problems in Mathematics.

Let $R$ be a commutative ring.

Then prove that $R$ is a field if and only if $\{0\}$ is a maximal ideal of $R$.
 

Proof.

$(\implies)$: If $R$ is a field, then $\{0\}$ is a maximal ideal

Suppose that $R$ is a field and let $I$ be a non zero ideal:
\[ \{0\} \subsetneq I \subset R.\]

Then the ideal $I$ contains a nonzero element $x \neq 0$. Since $R$ is a field, we have the inverse $x^{-1}\in R$.
Then it follows that $1=x^{-1}x \in I$ since $x$ is in the ideal $I$.

Since $1\in I$, any element $r \in R$ is in $I$ as $r=r\cdot 1 \in I$.
Thus we have $I=R$ and this proves that $\{0\}$ is a maximal ideal of $R$.

$(\impliedby)$: If $\{0\}$ is a maximal ideal, then $R$ is a field

Let us now suppose that $\{0\}$ is a maximal ideal of $R$.
Let $x$ be any nonzero element in $R$.

Then the ideal $(x)$ generated by the element $x$ properly contains the ideal $\{0\}$.
Since $\{0\}$ is a maximal ideal, we must have $(x)=R$.

Since $1\in R=(x)$, there exists $y\in R$ such that $1=xy$.
This implies that the element $x$ is invertible. Therefore any nonzero element of $R$ is invertible, and hence $R$ is a field.

The post Ring is a Filed if and only if the Zero Ideal is a Maximal Ideal first appeared on Problems in Mathematics.