Consider the left shift operator on the space of infinite sequences of complex numbers:
L(z1,z2,…)=(z2,z3,…).
Is L injective?
Yes <text> Yes</text> - incorrect No
Is L surjective?
Yes <text> Yes</text> - correct No
Find the eigenvalues and eigenvectors of L. Then complete the following sentence.
The complete set of eigenvalues of L is
0 <text> 0</text> - incorrect 1 0 and 1 all integers all real numbers all complex numbers
Let λ be an eigenvalue, the associated eigenvectors are :
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To determine whether the left shift operator L is injective, we need to check if different input sequences can produce the same output sequence. In this case, if two different sequences result in the same shifted sequence, then L is not injective.
To check if L is surjective, we need to determine if every possible output sequence can be obtained by applying the left shift operator to some input sequence.
To find the eigenvalues and eigenvectors of L, we need to solve the equation L(z1, z2, ...) = λ(z1, z2, ...), where λ represents the eigenvalue.
If we apply the left shift operator to the sequence (z1, z2, ...), we get (z2, z3, ...). In order for this sequence to equal λ times the original sequence (z1, z2, ...), we must have z2 = λz1, z3 = λz2, z4 = λz3, and so on. This gives us the recursive relationship zi = λzi-1.
From this recursive relationship, we can see that the eigenvalues λ must satisfy the equation:
λ * zi-1 = zi.
For example, if λ = 0, then zi-1 = 0, and all subsequent terms will also be zero. This means 0 is an eigenvalue, and the corresponding eigenvector is any sequence where all terms are zero except for possibly the first term.
For any other nonzero number λ, we can see that zi = λ^i * z0, where z0 is the initial value in the sequence. This means that λ raised to the power of the term index i gives the factor by which each term is scaled compared to the previous term.
Therefore, the complete set of eigenvalues of L is 0 and all complex numbers except 0. The associated eigenvectors depend on the chosen eigenvalue λ. If λ = 0, any sequence with all terms zero except the first term will be an eigenvector. If λ is any other complex number, the eigenvectors will be sequences of the form (z1, λz1, λ^2z1, ...), where z1 is a complex number.
Note: The answer choices for the eigenvalues in the question were not given, but the correct answer is "all complex numbers except 0".