Linear Algebra and Its Applications, Exercise 2.4.10

Exercise 2.4.10. Suppose the nullspace of a matrix is the set of all vectors x = \begin{bmatrix} x_1& x_2&x_3 \end{bmatrix}^T in \mathbf R^3 for which x_1 + 2x_2 + 4x_3 = 0. Find a 1 by 3 matrix A with this nullspace. Find a 3 by 3 matrix A' with the same nullspace.

Answer: If A = \begin{bmatrix} a_1&a_2&a_3 \end{bmatrix} and Ax = 0 then we have a_1x_1 + a_2x_2 + a_3x_3 = 0. We know that x_1 + 2x_2 + 4x_3 = 0 so one way to construct a suitable matrix A is to set a_1 = 1, a_2 = 2, and a_3 = 4 so that A = \begin{bmatrix} 1&2&4 \end{bmatrix}.

For a 3 by 3 matrix A', if A'x = 0  each row of A' times x is 0 so that a_{i1}x_1 + a_{i2}x_2 + a_{i3}x_3 = 0 for 1 \le i \le 3. Since x_1 + 2x_2 + 4x_3 = 0 we can construct a suitable matrix A' by setting each row of A' to \begin{bmatrix} 1&2&4 \end{bmatrix} or to any multiple of that vector. For example, one possible choice of A' is

A' = \begin{bmatrix} 1&2&4 \\ 2&4&8 \\ 4&8&16 \end{bmatrix}

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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2 Responses to Linear Algebra and Its Applications, Exercise 2.4.10

  1. Kenny says:

    Hi hecker, while we doing the A’ 3X3 matrix, since the information that x1+2×2+4×3=0 is given, can we be sure that A’ matrix is a rank1 matrix??

    • hecker says:

      A’ has rank 1 because the second and third columns are multiples of the first column, and thus are linearly dependent on the first column. The rank of a matrix is the number of linearly independent columns, and in this case there is only 1 such column.

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