Finding e^A for 2x2 Matrix: Issues with Det(A-(Lambda)I)

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The discussion focuses on calculating e^A for the 2x2 matrix A = [[1, 1], [-1, -1]]. The determinant calculation, det(A - λI), reveals that the eigenvalues are both zero, indicating that the matrix is singular. This confirms that the eigenvalues being zero is correct and not an error in the calculation. The participants suggest exploring the application of the matrix to a vector for further insights.

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I need to find e^A of this 2x2 matrix...A= 1 1
-1 -1

When I do det(A-(lambda)I) I get 0 for the eigenvalues, which makes no sense. Am I doing something wrong?
 
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Probably not. If the matrix is singular then at least one of the eigenvalues will be zero
Hmm...
[tex]det(A-I \lambda )=(-\lambda+1)(-\lambda-1)+1=\lambda^2[/tex]
So [itex]0[/itex] is a double root, and both of the eigenvalues are zero.

Why don't you see what happens if you apply this matrix to a vector?
 
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