Full rank of a matrix

  • #1
479
4

Homework Statement


Show that the matrix ##A## is of full rank if and only if ##ad-bc \neq 0## where $$A = \begin{bmatrix}
a & b \\
b & c
\end{bmatrix}$$

Homework Equations




The Attempt at a Solution


Suppose that the matrix ##A## is of full rank. That is, rank ##2##. Then by the rank-nullity theorem, the
dimension of the kernel is ##0##. This implies that there exists an inverse ##A^{-1}## but this will only occur if ##ad-bc \neq 0## otherwise our matrix ##A## will be singular. On the other hand, suppose ##ad-bc \neq 0##. Hence, ##A## is nonsingular and there exists an inverse ##A^{-1}## but this will occur only when the dimension of the kernel is ##0##, that is, of rank ##n = 2##.
 

Answers and Replies

  • #2
535
72
Your matrix has rank 2 iff the column vectors are linearly independent.
Can you show that the determinant of the matrix is zero iff the column vectors are linearly dependent ?
 
  • #3
479
4
Sorry $$A=\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$$ if that even matters!!!!

How could I do this without using anything on determinants???
 
  • #4
535
72
Show that ad - bc = 0 iff ##\vec u = (a,c) ## and ##\vec v = (b,d)## are linearly dependent
 
  • #5
479
4
Cheers!!!
 

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