Question

TRUE OR FALSE

The formula$\mathit{i}\mathit{m}\left(B\right)\mathbf{=}\mathit{i}\mathit{m}\left(B^{T}B\right)$ holds for all square matrices B.

Short answer

The solution is false.

Step-by-step solution

Given information

Consider the statement as follows:

“The formula $im\left(B\right)=im\left(B^{T}B\right)$holds for all matrices B.”

Explanation of the solution

Consider the matrix B as follows:

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

Then $im\left(B\right)$as follows:

$$\begin{gathered} im\left(B\right)=\left\{B\left[\begin{array}{c}x\\ y\end{array}\right]:\left[\begin{array}{c}x\\ y\end{array}\right]\in R^2\right\} \\ =\left\{\left|\begin{array}{cc}1& 0\\ 1& 0\end{array}\right|\left[\begin{array}{c}x\\ y\end{array}\right]:\left[\begin{array}{c}x\\ y\end{array}\right]\in R^2\right\} \\ =\left[\left[\begin{array}{c}x\\ y\end{array}\right]:x\in R^2\right] \end{gathered}$$

Now, simplify for $B^{T}B$as follows:

$$\begin{gathered} B^{T}B={\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]}^T\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right] \\ =\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right] \\ =\left[\begin{array}{cc}2& 0\\ 0& 0\end{array}\right] \end{gathered}$$

Now, simplify further as follows.

role="math" localid="1659614309579" $$\begin{gathered} im\left(B^{T}B\right)=\left\{B^{T}B\left[\begin{array}{c}x\\ y\end{array}\right]:\left[\begin{array}{c}x\\ y\end{array}\right]R^2\right\} \\ =\left\{\left[\begin{array}{c}2x\\ 0\end{array}\right]:x{R}^2\right\} \end{gathered}$$

It can be observed that, $\left[\begin{array}{c}1\\ 1\end{array}\right]\in im\left(B\right)$, but $\left[\begin{array}{c}1\\ 1\end{array}\right]\notin im\left(B^{T}B\right)$. Then,

$im\left(B\right)\notin im\left(B^{T}B\right)$

Hence, the given statement “The formula $im\left(B\right)=im\left(B^{T}B\right)$holds for all matrices B” is false.