32 Suppose I give you four nonzero vectors $r,n,c,l$ in $\mathbb{R}^2$.

(a) What are the conditions for those to be bases for the four fundamental subspaces $C(A^T), N(A), C(A), N(A^T)$ of a 2 by 2 matrix?

(b) What is one possible matrix $A$?

Here is the solution manual solution

We need $r^Tn=0$ and $c^Tl=0$. All possible examples have theform $acr^T$ with $a\neq 0$.

In my own solution I considered three cases based on the rank of $A$. My question is if this was unnecessary, and if the solution manual solution is actually a full/complete answer.

Consider the following matrix

$$A=\begin{bmatrix} 1&2\\ 2&5 \end{bmatrix}$$

This matrix has rank 2. The dimensions of the four fundamental subspaces are

$$\dim{C(A)}=\dim{C(A^T)}=2$$ $$\dim{N(A)}=\dim{N(A^T)}=0$$

We could have $r=n=(1,2)$, $c=l=(2,5)$. Then, $\{r,c\}$ would be a basis for both the column and row spaces. The nullspace and left nullspace have no vectors in their bases.

The vectors $(1,2)$ and $(2,5)$ are not perpendicular.

The second case is if $A$ has rank 1. In this case, we need one vector in the column space and one in the row space, and two additional vectors orthogonal to the first two. The vector that is orthogonal to the column space vector is a basis for the left nullspace, and the vector that is orthogonal to the row space is a basis for the nullspace. This seems to be similar to the case in the solution manual.

Was it ok for the solution manual to leave out the information that $r$, for example, should be in the row space and $n$ in the nullspace, and that $c$ should be in the column space and $l$ in the left nullspace?

The final case is for a rank 0 matrix $A$, which is just the zero matrix. The column and row spaces have no vectors in their basis. We need two independent vectors in the nullspace and two independent vectors in the left nullspace.

  • $\begingroup$ I'm surprised you chose to study linear algebra from Strang. Given your propensity for precision and rigor, it's absolutely a poor choice. He gives great intuition and talks very colloquially with the reader, but it is not a book for doing rigorous mathematics. $\endgroup$ Dec 27, 2022 at 19:55
  • $\begingroup$ @TedShifrin This is a prelude to tackling the book by Axler. I am taking the course 18.06 from MIT OCW, and right after I will take 18.700 which follows Axler. I like the intuition of the Strang course. He provides different intuitive angles, but the book is clearly not rigorous. There are no formal theorems in the book. $\endgroup$
    – xoux
    Dec 27, 2022 at 23:02

1 Answer 1


This book by Strang seems to be rather poorly written sometimes, and at the very least imprecise. I think when this problem named the vectors $r,n,c,l$ it was implied that they were vectors in the row space, null space, column space, and left nullspace, respectively.

If this is so, then we know the rank is 1. Therefore, since the column space and left nullspace are orthogonal complements, we need $c^Tl=0$.

Since the row space and the nullspace are orthogonal complements, we need $r^Tn=0$.

Thus, any matrix that has the form $cr^T$ has combinations of the vector $c$ in the columns and combinations of the vector $r$ in the rows. Rank is 1.


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