Chapter 4.1, Problem 1PS

To determine

Any 2 by 3 matrix of rank one, set the vector's according the given figure and discussing their orthogonal property.

Answer to Problem 1PS

The matrix A of order 2 by 3 be $A=\left[\begin{array}{l}\begin{array}{ccc}1& 2& 3\end{array}\\ \begin{array}{ccc}2& 4& 6\end{array}\end{array}\right]$is of rank 1

Explanation of Solution

Given:

Let the matrix A of order 2 by 3 be $A=\left[\begin{array}{l}\begin{array}{ccc}1& 2& 3\end{array}\\ \begin{array}{ccc}2& 4& 6\end{array}\end{array}\right]$is of rank 1, as 1^ndrow is obtained by multiplying $\frac{1}{2}$to 2^ndrow, So the number non-zero rows in echelon form of A is 1.

Hence, Rank of A is 1, which is equals to the dimension of row space.

From the figure, dimension of row space is r =1,

From rank-nullity theorem, dimension of null space of A is $n-r=3-1=2$, where n=3. Which must be orthogonal to the row space of A.

Since row space spanned by (1, 2, 3), now we have to choose null space such that their dot product with (1, 2, 3) is zero. Which is the condition for orthogonal.

So, null space $A=\left(2,-1,0\right),\left(0,-3,2\right)$

Now the similar way if we look at the dimension of column space of A is also 1, because the rank of A^Tis also 1.

Which implies, Null space of A^Tis $m-r=2-1=1$, where m=2.

We choose null space of A^Tis (2,-1). Since the column space vector is orthogonal (2,-1).

Now, Let us set a figure with above information.