If a Matrix $A$ is Singular, then Exists Nonzero $B$ such that $AB$ is the Zero Matrix

Ohio State University exam problems and solutions in mathematics

Problem 301

Let $A$ be a $3\times 3$ singular matrix.

Then show that there exists a nonzero $3\times 3$ matrix $B$ such that
\[AB=O,\] where $O$ is the $3\times 3$ zero matrix.

 
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Proof.

Since $A$ is singular, the equation $A\mathbf{x}=\mathbf{0}$ has a nonzero solution.
Let $\mathbf{x}_1$ be a nonzero solution of $A\mathbf{x}=\mathbf{0}$.
We define the $3\times 3$ matrix $B$ to be
\[B=[\mathbf{x}_1, \mathbf{0}, \mathbf{0}],\] that is, the first column of $B$ is the vector $x_1$ and the second and the third column vectors are $3$-dimensional zero vectors.

Then since $x_1\neq \mathbf{0}$, the matrix $B$ is not the zero matrix.
We have
\begin{align*}
AB&=A[\mathbf{x}_1, \mathbf{0}, \mathbf{0}]\\
&=[A\mathbf{x}_1, A\mathbf{0}, A\mathbf{0}]\\
&=[\mathbf{0}, \mathbf{0}, \mathbf{0}]=O,
\end{align*}
since $A\mathbf{x}_1=\mathbf{0}$.

Thus we have found the nonzero matrix $B$ such that the product $AB=O$.

Comment.

This is one of the midterm exam 1 problems of linear algebra (Math 2568) at the Ohio State University.

In the exam the following hint was given:

Hint: Let $B=[\mathbf{x}_1, \mathbf{x}_2, \mathbf{x}_3]$, where $\mathbf{x}_i$ is the $i$-th column vector of $B$ for $i=1,2,3$. Then $AB=[A\mathbf{x}_1, A\mathbf{x}_2, A\mathbf{x}_3]$.

Common Mistake

Several students wrote “Since $A$ is singular, $A$ has a solution”.
This does not make sense. A solution of what? You need to remember the definition correctly.

A matrix $A$ is singular if the equation $A\mathbf{x}=\mathbf{0}$ has a nonzero solution $\mathbf{x}$.

Also, note that $\mathbf{x}$ is a vector, not a number.

Midterm 1 problems and solutions

The following list is the problems and solutions/proofs of midterm exam 1 of linear algebra at the Ohio State University in Spring 2017.

  1. Problem 1 and its solution: Possibilities for the solution set of a system of linear equations
  2. Problem 2 and its solution: The vector form of the general solution of a system
  3. Problem 3 and its solution: Matrix operations (transpose and inverse matrices)
  4. Problem 4 and its solution: Linear combination
  5. Problem 5 and its solution: Inverse matrix
  6. Problem 6 and its solution: Nonsingular matrix satisfying a relation
  7. Problem 7 and its solution: Solve a system by the inverse matrix
  8. Problem 8 and its solution (The current page): A proof problem about nonsingular matrix

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9 Responses

  1. Michael Braun says:

    There is an error. The problem says A is singular. The proof says A is nonsingular.

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