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?

Answer: The answer is no, as shown by the following counterexample:

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.

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