This statement is true/false? Justify your answer. If #A^k=0# for a square matrix #A#, then all the eigenvalues of #A# are zero.

1 Answer
Feb 7, 2018

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# \mbox{The statement is True.} #

Explanation:

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# \mbox{1) Let:} \qquad \lambda \ \ \mbox{be an eigenvalue of} \ \ A. #

# \mbox{2) Then, by definition:} \qquad A v = \lambda v, \ \mbox{for some non-zero vector} \ v. #

# \mbox{3) Then by (2), and as} \ \lambda \ \mbox{is a scalar, we note:} #

# \qquad \qquad \qquad A^2 v = A(Av) = A ( \lambda v ) = \lambda (A v) = \lambda ( \lambda v ) = \lambda^2 v; #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad :. \qquad \qquad A^2 v = \lambda^2 v. #

# \mbox{4) Then by (3), and then by (2), and as} \ \lambda \ \mbox{is a scalar, we note:} #

# \qquad \qquad \qquad A^3 v = A(A^2 v) = A ( \lambda^2 v ) = \lambda^2 (A v) = \lambda^2 ( \lambda v ) = \lambda^3 v; #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad :. \qquad \qquad A^3 v = \lambda^3 v. #

# \mbox{5) Continuing, we see [or by an elementary induction]:} #

# A^n v = A(A^{n-1} v) = A ( \lambda^{n-1} v ) = \lambda^{n-1} (A v) = \lambda^{n-1} ( \lambda v ) #

# \qquad \quad \ = \lambda^n v, \qquad \mbox{for any positive integer} \ n; #

# \qquad \qquad \qquad \qquad \qquad :. \quad \qquad A^n v = \lambda^n v, \qquad \quad \mbox{for any positive integer} \ n. #

# \mbox{6) Now use the conclusion in (5), with} \quad n = k, #

# \qquad \qquad \quad \mbox{where} \ k \quad \mbox{is from the given about} \ \ A. #

# \qquad \qquad \qquad \qquad \qquad :. \quad \qquad A^k v = \lambda^k v. #

# \qquad \qquad \qquad \qquad \qquad :. \quad \qquad 0 v = \lambda^k v. #

# \qquad \qquad \qquad \qquad \qquad :. \quad \qquad \lambda^k v = 0. #

# \mbox{7) Now recall by (2):} \quad \v \ \mbox{is a non-zero vector. Thus, at least one} \ \ \mbox{entry in} \ v \ \mbox{is non-zero. Let this non-zero entry be} \ a. \mbox{Hence,} \ \mbox{using the conclusion in (6):} #

# \qquad \qquad \qquad \qquad \qquad :. \quad \qquad \lambda^k a = 0, \qquad \mbox{and} \ \ a \ \ \mbox{is non-zero}. #

# \qquad \qquad \qquad \qquad \qquad:. \quad \qquad \lambda^k = 0. #

# \qquad \qquad \qquad \qquad \qquad:. \quad \qquad \lambda = 0. #

# \qquad \quad \ \ :. \quad \qquad \mbox{any eigenvalue of} \ \ A \ \ \mbox{is 0}. \qquad \mbox{QED} #