Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $A$ is a $2\times2$ matrix. How do I prove that, if $\det(A) < 0$, then $A$ is a diagonalizable matrix over $\mathbb{R}$?

share|improve this question
7  
You can't, because it's false. –  Chris Eagle May 3 '12 at 21:10
1  
The Jordan form must actually be diagonal, since a $2x2$ Jordan block has nonnegative determinant. –  Grumpy Parsnip May 3 '12 at 21:17
4  
@ChrisEagle's comment was made before the OP was modified to include the 2x2 hypothesis. –  Grumpy Parsnip May 3 '12 at 21:18
2  
@JimConant : If you write 2x2 in $\TeX$, with the letter $x$, then it looks like an $x$, not like a $\times$. If you write 2\times 2, then it looks like $2\times 2$. –  Michael Hardy May 3 '12 at 21:53
    
@MichaelHardy: Yes I know, I was just being lazy. –  Grumpy Parsnip May 4 '12 at 1:03
add comment

1 Answer 1

up vote 14 down vote accepted

Since the matrix is $2 \times 2$, its characteristic polynomial is given by $x^2 - \operatorname{tr} A \cdot x + \det A$. Since $(\operatorname{tr} A)^2 - 4 \det A > 0$ by $\det A < 0$, this polynomial has two distinct real zeroes, i.e., $A$ has two distinct real eigenvalues. Since each eigenvalue has at least one eigenvector, the geometric and algebraic multiplicities of the eigenvalues coincide, i.e., $A$ is diagonalizable.

share|improve this answer
3  
Also we can say that product of eigen values is the determinant of the matrix. A is $2\times2$ matrix with $\det(A)<0$ that means $\lambda_{1}\lambda_{2}<0$ , where $\lambda$ refers to the eigen values of given matrix $A$.Consequently $\lambda_{i}$ ,i=1,2 must be of opposite signs or we can say that A must have two distinct eigen values having oppsite signs. Which results in $A$ is diagonalizable. –  srijan May 12 '12 at 12:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.