Does anyone know how to prove that eigenvalues (including multiplicities) of a triangular matrix coincides with its diagonal entries? Thank you so much!
Linear Algebra: Eigenvalue Proof Question
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linear-algebra
eigenvalues-eigenvectors
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0There is a proof in Axler for instance. It avoids determinants. – 2017-02-13
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1Hint: what is the determinant of a diagonal matrix? – 2017-02-13
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0@lhf thanks so much, do you know where I could find that proof? – 2017-02-13
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0You can take a look at this lesson: http://lem.ma/vC You will learn something from each of the videos in that lesson, but the very last video deals with a triangular matrix. – 2017-02-13
1 Answers
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All diagonal entries are eigenvalues
Proof by induction on the order $n$ of the matrix. $n=1$ is easy.
For $n>1$, by induction, $a_{11}, a_{22}, \dots a_{n-1,n-1}$ are eigenvalues of $A$. Now the last row of $A-a_{nn}I$ is all zeros and so $A-a_{nn}I$ does not full rank. This implies that the kernel of $A-a_{nn}I$ is not trivial and so $a_{nn}$ is an eigenvalue.
All eigenvalues appear in the diagonal
If $\lambda$ is does not appear in the diagonal, then $B=A-\lambda I$ does not have any zero in the diagonal and so $Bx=b$ can be solved for all $b$. This means that $B$ is invertible and so $B$ is injective. In other words, the kernel of $B=A-\lambda I$ is $0$ and $\lambda$ is not an eigenvalue of $A$.