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Why the following command prints Smith but not \Smith?

echo \Smith
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The backslash is an escape character that:

shall preserve the literal value of the following character, with the exception of a <newline>. ... The <backslash> ... shall be removed

So \S means the same thing as S, because S is not a newline character and also not a shell special character that could be escaped ($, ", ', {, [, `, \, |, &, ;, <, >, (, ), ?, *, [, #, ~, =, %, , or tab). To include a literal backslash, escape it in turn:

echo \\Smith

note: this is the specified behavior for your example unquoted case, but it can change if it is otherwise quoted or read in from a file

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\S escapes the S, which is not a special character, thus gives S. You need to double the backslash to print it: echo \\Smith

EDIT: But more generally, it's better to use printf. See the difference between echo x\\by, which outputs "y" (the "x" gets overwritten by the backspace \b) with some versions of echo (dash, zsh), and printf "%s\n" x\\by, which outputs "x\by".

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