Tricky Matrices

Algebra Level pending

Two non null matrices $A$ and $B$ satisfy $AB=B$ and $BA=A$ . Then the matrix $A^2+B^2$ equals

Here $I$ stands for the identity matrix

AB

A+B

I

3 solutions

Chris Lewis
Feb 27, 2020

We have $A^2 = BA\cdot BA = B\cdot AB \cdot A = B \cdot B \cdot A = B \cdot BA = B \cdot A = BA = A$

and

$B^2 = AB\cdot AB = A\cdot BA \cdot B = A \cdot A \cdot B = A \cdot AB = A \cdot B = AB = B$

so $A^2+B^2=A+B$ .

That's you 👶

A Former Brilliant Member - 1 year, 3 months ago

Karan Chatrath
Feb 25, 2020

Let the matrix $A$ have $m$ rows and $n$ columns and matrix $B$ have $n$ rows and $p$ columns. The matrix $AB$ will therefore have $m$ rows and $p$ columns. The number of columns of $A$ and the number of rows of $B$ must be same, otherwise, the multiplication $AB$ is not possible.

Since:

$AB = B$

We can conclude that

$m = n$

We can make similar arguments and conclude that

$m = n = p$

Therefore $A$ and $B$ are square matrices. Consider the expression:

$X = A^2 + B^2$ $\implies X = A \cdot A + B \cdot B$

Assuming the matrices $A$ and $B$ are invertible, from the given information we get:

$A = B B^{-1} = I$ $B = A A^{-1} = I$

This leads to

$X = A \cdot A + B \cdot B = A\left(B B^{-1}\right) + B\left(A A^{-1}\right)$

Which simplifies to:

$\boxed{X = A^2 + B^2 = A + B =2I}$

Is there a way of proving this expression without assuming that $A$ and $B$ are invertible? I am curious. If not, a note that the matrices are full-rank (row or column) should be mentioned in the problem statement.

Karan Chatrath - 1 year, 3 months ago

What I have done is as follows

$AB=B\implies AB=IB\implies (A-I) B=\varnothing\implies A=I$ . Similarly, $BA=A\implies B=I$ . Hence $A^2=A, B^2=B$ , and $A^2+B^2=A+B$ .

Even though it is inherent that matrices $A$ and $B$ are invertible, it was not necessary to use this fact explicitly.

A Former Brilliant Member - 1 year, 3 months ago

The matrices $A=\begin{pmatrix} 1 & 2\\ 0 & 0 \end{pmatrix}$

and

$B=\begin{pmatrix} 1 & -2\\ 0 & 0 \end{pmatrix}$

satisfy the condition, but $A\neq I$ and neither matrix is invertible...

Chris Lewis - 1 year, 3 months ago

@Chris Lewis – Brilliant ! In fact the matrices

$A=\left[{\begin {array} {cc} 1 & a\\ 0 & 0\\ \end {array} }\right ]$

and

$B=\left[{\begin {array} {cc} 1 & b\\0 & 0\\ \end {array}}\right ]$

are non invertible non identity matrices satisfying those conditions.

A Former Brilliant Member - 1 year, 3 months ago

Aha - found a way!

Chris Lewis - 1 year, 3 months ago

Rocco Dalto
Apr 4, 2020

Not assuming $A$ and $B$ are invertible.

$AB = B \implies (AB)A = BA = A \implies A(BA) = A \implies A^2 = A$

and

$BA = A \implies (BA)B = AB = B \implies B(AB) = B \implies B^2 = B$

$\implies A^2 + B^2 = A + B$ .

Tricky Matrices

3 solutions

0 pending reports