Linear Algebra and Its Applications, Exercise 3.3.8

Exercise 3.3.8. Suppose that P is a projection matrix from \mathbb R^n onto a subspace S with dimension k. What is the column space of P? What is its rank?

Answer: Suppose that b is a arbitrary vector in \mathbb R^n. From the definition of P we know that Pb is a vector in S. But Pb is a linear combination of the columns of P, so that Pb also is in the column space \mathcal{R}(P). Since any vector in S can be expressed as Pb for some b,  all vectors in S are also in \mathcal{R}(P), so that S \subseteq \mathcal{R}(P).

Now suppose that v is an arbitrary vector in the column space \mathcal{R}(P). Then v can be expressed as a linear combination of the columns of P for some set of coefficients a_1, a_2, \dots, a_n. Consider the vector w = (a_1, a_2, \dots, a_n). We then have v = Pw by the definition of w. But if  v = Pw for some w then v is in S. So all vectors in \mathcal{R}(P) are also in S and thus \mathcal{R}(P) \subseteq S.

Since S \subseteq \mathcal{R}(P) and \mathcal{R}(P) \subseteq S we then have S = \mathcal{R}(P): The column space of P is S.

The rank of P is the dimension of its column space. But since S is the column space of P, the rank of P is k, the dimension of S.

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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