## Linear Algebra and Its Applications, Exercise 2.4.21

Exercise 2.4.21. Suppose that for two matrices $A$ and $B$ the associated subspaces (column space, row space, null space, and left nullspace) are the same. Does this imply that $A = B$? $A = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} \qquad B = \begin{bmatrix} -1&0 \\ 0&1 \end{bmatrix}$
We have the column spaces $\mathcal R(A) = \mathcal R(B) = \mathbf R^2$. We also have the row spaces $\mathcal R(A^T) = \mathcal R(B^T) = \mathbf R^2$. Since the matrices are nonsingular the nullspaces $\mathcal N(A)$ and $\mathcal N(B)$ contain only the zero vector and are thus equal. For the same reason the left nullspaces $\mathcal N(A^T)$ and $\mathcal N(B^T)$ also contain only the zero vector and are also equal.
Thus equality of the fundamental subspaces of $A$ and $B$ does not imply that $A = B$. Also note that we cannot even conclude that $A = cB$ for some scalar $c$, as shown by the counterexample above.
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .