The equation Tu = u is equivalent to (T I)u =0, so is an eigenvalue of T if and

Question

The equation Tu = u is equivalent to (T I)u =0, so is an eigenvalue of T if and only if T Iis not injective (one to one).
What does this mean?
What does this mean?

Explanation / Answer

If for a non-zero vector u, Tu=u for some constant ,------(*) then we can rewrite that as Tu-u=0. Write u=Iu (where I is the identity matrix), so thatthe above can be written as Tu-Iu=0 This can be rewritten as (T-I)u=0. ---------------(**) Thus, if (*) holds then it is equivalent to (**). Note that if we write A = T-I, then this is equivalent towriting Au=0 => u is in Ker(A), the kernel of A. Since u isnonzero by assumption, it follows that the kernel of A is not zero.But for any linear transformation A, it is injective (1-1) if andonly if its kernel is non-zero. Hence, to say that isan eigenvalue of a linear map T, is equivalent to saying thatT-I is not injective. Does this make sense now?

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